**1 . ప్రధాన సంఖ్యలు మరియు సంయుక్త సంఖ్యలు (Prime and Composite numbers)**

**2. భాజనీయత సూత్రాలు (Test of divisibility Rules)**

**3. సంఖ్యల రకాలు (types of numbers)**

**4. వాస్తవ సంఖ్యలు (Real numbers)**

**5. భిన్నాలు మరియు దశాంశ భిన్నాలు (Fractions and Decimal Fractions)**

**6. క .సా .గు . మరియు గ . సా .భా. యూక్లిడ్ భాగహార పద్ధతి (LCM and HCF – Euclid division Lemma)**

**7. వర్గాలు – వర్గ మూలాలు (Squares – Square roots)**

**8. ఘనాలు – ఘన మూలాలు ****(****Cubs**** – ****Cube**** roots**

**9****. సంఖ్యల అమరిక మరియు సంఖ్యల ఫజిల్ ****(****Numbers Pattern and ****Numbers puzzle****)**

**10****. యూక్లిడ్ భాజనీయత సూత్రం **** ****(****Euclid Division lemma)**

**11****.** **సంవర్గమానము ****(****Logarithms****)**

TS 6th Class Maths ConceptTS 10th class maths concept (E/M)Ts Inter Maths IA ConceptTS 10th Class Maths Concept (T/M)TS 10th Class Maths Concept (T/M)

**1.** **నిష్పత్తి – అనుపాతం **** (****Ratio**** and Proportions****)**

**2****.** **సాదారణ వడ్డీ – బారు వడ్డీ ****(****Simple Interest – Compound Interest)**

**3. కాలము – దూరం ****(Time and Distance****)**

**4****.** **కాలము – పని ****(****Time and Work)**

**5****.** **లాభము – నష్టము – డిస్కౌంట్ ****(****Profit and Loss – Discount)**

**6****.** **పన్నులు ****(****Taxes)**

**1****. సమితి బావన – సమితి భాష ****(****Concept of set**** – ****Set language****)**

**2. శూన్య సమితి **** ****(****Empty set****)**

**3. ****ఉప సమితి **** ****(****Sub set****)**

**4****.** **పరిమిత – అపరిమిత సమితులు **** (****Finite and Infinite Sets****)**

**5****.** **సమ సమితులు **** **** ****(****Equality of sets)**

**6.** **కార్డినల్ సంఖ్య ****(****Cardinal number of a set****)**

**7****.** **సమితుల పరిక్రియలు ****(Set operations****)**

**8****. వెన్ చిత్రాలు **** ****(****Venn diagrams****)**

**1****. బీజ గణిత పరిచయం ****(****Introduction to Algebra****)**

**2. ఘాతాలు – ఘతాంకాలు **** ****(****Exponents – Powers****)**

**3. ****ప్రత్యేక లబ్ధాలు ****(****Special Products****)**

**4****.** **బహుపదులు **** **** ****(****Polynomials****)**

**5****.** **రేఖీయ సమాసాలు , వాటి గ్రాఫ్లు ****(****Linear equations , their graphs)**

**6****. ****వర్గ సమీకరణాలు **** ****(Quadratic Equations****)**

**7.** **శ్రేడులు ****(****Progressions)**

**1****. రేఖా గణిత పరిచయం ****(****Introduction to Geometry****)**

**2. జ్యామితి అభివృద్ధిలో భారతదేశం యొక్క సహకారం**

** ****(C****ontribution**** of India in the development of geometry ****)**

**3. ****యూక్లిడ్ రేఖా గణితం ****(****Euclid Geometry****)**

**4****.** **సరళ రేఖలు మరియు కోణాలు **** ****(****Lines and Angles****)**

**5****.** **సరూప త్రిభుజాలు ****(Similar Triangles****)**

**6****. **** సర్వ సమాన త్రిభుజాలు **** ****(Congruent Triangles****)**

**7.** **పైథాగరస్ సిద్దాంతం ****(****Pythagoras Theorem ****)**

**8****. వృత్తాలు – వృత్త ధర్మాలు ****(****Circles – Properties of Circles****)**

**9****.** **వృత్తాల నిర్మాణం **** **** ****(****Construction of circles** **)**

**10****. ****త్రిభుజాలు ****(Triangles****)**

**11****.** ** బహుభుజులు – చతుర్భుజాలు **** ****(****Polygons – ****Quadrilaterals** **)**

**12****.** **నిరూపక జ్యామితి ****(****Coordinate ****Geometry****)**

**13. **** సరళ రేఖలు **** ****(****Straight lines****)**

**1. పరిధి మరియు వైశాల్యం : ( Perimeter and Area )**

** ****త్రిభుజం(triangle ),చతురస్రం(s quare ), దీర్ఘ చతురస్రం (rectangle ), వృత్తం (circle ), కంకణం(ring ) **** **

**2. ఘనము – దీర్ఘ ఘనము ****( Cube **** – **** ****Cuboid****)**

**3. ****స్తూపం ****(Cylindre****)**

**4****.** ** శంఖువు ****(Cone****)**

**5****.** **గోళము – అర్థ గోళము ****(Sphere – Hemi Sphere****)**

**1. దత్తాంశ సేకరణ – వర్గీ కరణ ****( ****Collection of data – Classification of data**** )**

**2. ****పౌనఃపున్య పట్టిక ****(****Frequency Table****)**

**3****.** ** గణన చిహ్నాలు ****(****Tally Marks** **)**

**4****.**** ****ప** **ట చిత్రాలు ****(****Graphs****)**

**5. కేంద్రీయ స్థాన కొలతలు (Measures of Central Tendency)**

** అంక గణితం (Arithmetic mean ), మధ్య గతం (Median), బాహులకం ( Mode )**

**6****.**** ****సంభావ్యత ****(****Probability ****)**

**1. త్రికోణమితి పరిచయం ( Introduction of Trigonometry )**

**2. ****త్రికోనమితీయ నిష్పత్తులు ****(****Trigonometric Ratios****)**

**3****.** ** ప్రత్యేక కోణాల త్రికోనమితీయ నిష్పత్తులు ****(****Specific Trigonometric ratios of angles** **)**

**4****.**** ****పూరక కోణాల ****త్రికోనమితీయ నిష్పత్తులు** **(Trigonometric ratios of ****complimentary ****angles** **)**

**5. త్రికోనమితీయ సర్వ సమీకరణాలు **** ****(****Trigonometric Identities****)**

**6****.** **త్రికోణమితి అనువర్తనాలు ****( Applications **** of Trigonometry**** )**

]]>

These Model papers 2022 to do help the intermediate First-year Maths students.

TS 10th class maths concept (E/M)

TS 10th Class Maths Concept (T/M)

TS 10th Class Maths Concept (T/M)

TS Inter Maths 1A Model papers 2022 as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Model papers 2022 to do help the intermediate First-year Maths students.

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TS Inter Maths 1B Model papers 2022 as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Model papers 2022 to do help the intermediate First-year Maths students.

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TS Inter 2nd Year Maths 2A Concept

TS Inter 2nd Year Maths 2B Concept

TS 10th class maths concept (E/M)

TS 10th Class Maths Concept (T/M)

TS 10th Class Maths Concept (T/M)

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A) ₹ 200

B) ₹ 600

C) ₹ 500

D) ₹ 300

` `

A) ₹ 7865

B) ₹ 12875

C) ₹875

D) ₹13865

A) ₹ 78650

B) ₹ 2350

C) ₹12350

D) ₹13050

** Sol: we have to insert 4 rational numbers between 1 and and 1**

** 4 +1 = 5 **

**1 can be written as **

**Lcm of (4, 5) = 20**

**2. The prime factorisation of a natural number (n) is 2 ^{3} × 3^{2} × 5^{2} × 7. How many consecutive zeros will it have at the end of it? Justify your answer.**

**Sol: **

** n = 2 ^{3} × 3^{2} × 5^{2} × 7 **

** = 2 × 2 ^{2} × 3^{2} × 5^{2} × 7 **

** = 2 × 3 ^{2} × 7 (5^{2} × 2^{2}) **

** = 2 × 3 ^{2} × 7 × (10)^{2} = 12600**

**∴ n has 2 consecutive zeros at the end**

**3. Find the value of **

**Sol:**

**4. Write any two irrational numbers between 3 and 4**

**Sol: **

** **

** 5. Find the value of **

** **

** 6. Find the HCF and LCM of 90, 144 by using the prime factorisation method**

**Sol:**

**7. Is rational or irrational? justify your answer**

**Sol:**

**8. Expand **

**Sol:**

**9. Find the HCF of 24 and 33 by using division method**

**Sol:**

**33 > 24**

** 33 = 24 × 1 + 9**

** 24 = 9 × 2 + 6**

** 9 = 6 × 1 + 3**

** 6 = 3 × 2 + 0**

**∴ HCF of 24 ****and 33 = 3**

**10. Find the value of **

**Sol:**

**11. Ramu says, “if = 0, the value of x is 0”. Do you agree with him? Give reason.**

**Sol:**

**Sol: let 10, 12, 16 be three two-digit numbers**

** Prime factarisation of 10 = 2 × 5 = 2 ^{1}× 5^{1}**

** Prime factarisation of 12 = 2 × 2× 3 = 2 ^{2} × 3**

** Prime factarisation of 16 = 2 × 2× 2× 2 = 2 ^{4}**

**HCF of 10, 12 and 16 = 2 ^{1} = 2**

**LCM of 10, 12 and 16 = 2 ^{4} × 3× 5 = 8 × 3× 5 = 120**

**2. Give an example for each of the following**

**(i) The product of two irrational numbers is a rational number**

**(ii) The product of two irrational numbers is an irrational number**

**Sol:**

** **

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An equation of the form ax + by + c = 0 where a, b, c are real numbers and a^{2} + b^{2} ≠ 0 is called a linear equation in two variables x and y.

Two linear equations in the same variables are called a pair of linear equations in two variables

a_{1}x + b_{1}y + c_{1} = 0; a_{2}x + b_{2}y + c_{2} = 0 are the pair of linear equations in two variables in x and y.

For linear equations in two variables, there are infinitely many solutions.

Ex: x + y = 10

x =7, y = 3; x = 6, y = 4; x = 1, y = 9; x =2, y = 8; x=3, y = 7 like we have infinitely many solutions.

⇰ For finding exact values of x and y we have to know two linear equations.

⇰ A pair linear equations in two variables solved by four methods

- Graphical method
- Substitution method
- Elimination method
- Cross – Multiplication method

**Solving the pair of linear equations in two variables by using graphical Method: **

After plotting the points in the above tables in Cartesian plane, we observe that two straight lines intersect at the point (2, 1)

There is only one solution for this pair of linear equations in two variables.

These equations are known as consistent pair of linear equations and they have a unique solution.

After plotting the points in the above tables in Cartesian plane, we observe that two straight lines are parallel

There is no solution for this pair of linear equations in two variables.

These equations are known as inconsistent pair of linear equations and they have no solution.

After plotting the points in the above tables in Cartesian plane, we observe that two straight lines are coincide

There are infinitely many solutions for this pair of linear equations in two variables.

These equations are known as consistent pair of linear equations and they have infinitely solution.

**Consistent and inconsistent:**

If the system of equations has a solution, then they are consistent. If the system of equations has no solution, then they are inconsistent.

**The relationship between coefficients and the nature of the equation system:**

**1. Draw the graph of the following pair of linear equations in two variables and find their solution from the graph**

** 3x – 2y = 2 and 2x + y = 6**

The two lines intersect at the point (2, 2)

∴ solutions is x = 2, y = 2

**2. Draw the graph of the following pair of linear equations in two variables and find their solution from the graph**

** x – 2y = –1 and 2x – y – 4 = 0**

The two lines intersect at the point (3, 2)

∴ solutions is x = 3, y = 2

**3. Represent the solution of linear equation graphically**

** x – 2y = –3 and 2x + y = 4**

The two lines intersect at the point (1, 2)

∴ solutions is x = 1, y = 2

**1. Neha went to a sale to purchase some points and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts is two less than twice the number of points purchased. Also, the number of number of skirts is four less than four times the number of pants purchased”.**

**Help her friend to find how many pants and skirts Neha bought.**

**Sol:**

Let the number of points = x

and the number of skirts = y

the number of skirts is two less than twice the number points purchased

⟹ y = 2x – 2

the number of number of skirts is four less than four times the number of pants purchased

⟹ y = 4x – 4

The two lines intersect at the point (1, 0)

∴ solution is x = 1, y = 2

No. of Points = 1 and no. of skirts = 0

**2. 10 students of a class X took part in a Mathematics quiz. If the no. of girls is 4 more than the no. of boys, then find the no, of boys and no. of girls who took part in the quiz.**

**Sol:**

let the no. of boys = x

No. of girls = y

Total no. of students = 10

⟹ x + y = 10

The no. of girls is 4 more than the no. of boys

⟹ y = x + 4

⟹ x – y = – 4

The two lines intersect at the point (3, 7)

∴ solution is x = 3, y = 7

No. of boys = 3 and no. of girls = 7

**3. Half the perimeter of a rectangular garden, whose length is 4m more than its width is 36 m. Find the dimensions of the garden.**

**Sol:**

let the width of garden = x m.

Length of garden = y m.

Length of garden is 4m mote than its width

⟹ y = x + 4

x – y = – 4

half of the perimeter of rectangular garden is 36 m.

⟹ x + y = 36

⟹ x – y = – 4

The two lines intersect at the point (16, 20)

∴ solution is x = 16, y = 20

Width of garden = 16 m and Length of garden = 20 m

**4. The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will be increased by 50 sq units. Find the length and breadth of rectangle.**

**Sol:**

Let the length of rectangle = x units

the breadth of rectangle = y units

The area of rectangle = xy sq units

If the Length of rectangle is reduced by 5 units and breadth is increased by 2 units, then area will be reduced by 80 sq units.

⟹ (x – 5) (y + 2) = xy – 80

⟹ xy + 2x – 5y – 10 = xy – 80

⟹ 2x – 5y = – 70

If length is increased by 10 and breadth decreased by 5 then area will be increased by 50 squnits

⟹ (x + 10) (y – 5) = xy + 50

⟹ xy – 5x + 10y – 50 = xy + 50

⟹ – 5x + 10y = 100

The two lines intersect at the point (40, 30)

∴ solution is x = 40, y = 30

Length of rectangle = 40 units and Breadth of rectangle = 30 units

**5. In X class, if three students sit on each bench one student will be left, if four students sit on each bench, one bench will be left. Find the number of students and number of benches in that class.**

**Sol:**

Let the no. of students = x

The no. of benches = y

If three students sit on each bench one student will be left

⟹ 3y = x – 1

⟹ x – 3y = 1

If four students sit on each bench one bench will be left

⟹ 4 (y – 1) = x ⟹ 4y – 4 = x

⟹ x – 4y = – 4

The two lines intersect at the point (16, 5)

∴ solution is x = 16, y = 5

No. of students = 16 and No. of benches = 5

**The relationship between coefficients and the nature of the equation system Related Problems:**

**1. Check whether the following pair of linear equations have a unique solution, infinitely many solutions or no solution**

**(i) 2x + 3y = 1; 3x – y = 7**

**(ii) 2x + 3y = 5; 4x + 6y = 10**

**(iii) 5x + 3y = 4;10x + 6y = 12**

Sol:

**(i)** Given equations are x + 3y = 1; 3x – y = 7

a_{1} = 1, b_{1} = 3, c_{1} = 1; a_{2} = 3, b_{2} = –1, c_{2} = 7

∴ Given equations have unique solution

**(ii)** Given equations are 2x + 3y = 5; 4x + 6y = 10

a_{1} =2, b_{1} = 3, c_{1} = 5; a_{2} = 4, b_{2} = 6, c_{2} = 10

∴ Given equations have infinite solutions

**(iii) **Given equations are 5x + 3y = 4;10x + 6y = 12

a_{1} =5, b_{1} = 3, c_{1} = 4; a_{2} = 10, b_{2} = 6, c_{2} = 12

∴ Given equations have no solution

**2. For what value of ‘p’ the following pair of equations has a unique solution**

** 2x + py = – 5 and 3x + 3y = – 6**

Sol:

Given equations are 2x + py = – 5, 3x + 3y = – 6

a_{1} =2, b_{1} = p, c_{1} = –5; a_{2} = 3, b_{2} = 3, c_{2} = –6

Above equations have a unique solution

∴ Given pair of linear equations have a unique solution when p ≠ 2

**3. For what value of ‘p’ the following pair of equations has a unique solution**

** px + 3y –(p – 3)= 0 and 12x + py – p = 0**

Sol:

Given equations are px + 3y – (p – 3) = 0, 12x + py – p = 0

a_{1} =p, b_{1} = 3, c_{1} = – (p – 3),; a_{2} =12, b_{2} = p, c_{2} = –p

Above equations have infinitely many solution

∴ Given pair of linear equations have infinitely many solutions when p = ±6

**4. For what value of ‘k’ the following pair of equations represent coincident lines.**

** 3x + 4y + 2 = 0 and 9x + 12y + k = 0**

Sol:

Given equations are 3x + 4y + 2 = 0 and 9x + 12y + k = 0

a_{1} =3, b_{1} = 4, c_{1} = 2; a_{2} = 9, b_{2} = 12, c_{2} = k

Above equations are coincident lines

∴ Given pair of linear equations are coincident when k = 6

**5. For what value of ‘k’ the following pair of equations represents the parallel lines**

** 2x – ky + 3 = 0 and 4x + 6y – 5 = 0**

Sol:

Given equations are 2x – ky + 3 = 0, 4x + 6y – 5 = 0

a_{1} =2, b_{1} = –k, c_{1} = 3; a_{2} = 4, b_{2} = 6, c_{2} = –5

Above equations represents parallel lines

∴ Given pair of linear equations represents parallel lines when k = – 3

In this method, we make one of the variables x or y as the subject from the 1^{st} equation (or 2^{nd} equation). Substitute this equation in 2^{nd} equation (or 1^{st} equation) and get the value of the variable involved, then by substituting this value in any one of the equations we get the value of second variable.

** **

**1. Solve the following pair of equations by using substitution method**

** 2x – y = 5 and 3x + 2y = 11**

Sol:

Given equations are 2x – y = 5 ———— (1)

3x + 2y = 11———– (2)

From (1)

y = 2x – 5

substitute y value in (2)

⟹ 3x + 2(2x – 5) = 11

3x + 4x – 10 = 11

7x = 11 + 10 = 21

x = 3

now y = 2x – 5

⟹ y = 2(3) – 5 = 6 – 5 = 1

∴ the solution is x = 3 and y = 1

**2. Solve the following pair of equations by using substitution method**

** 2x + 3y = 9 and 3x + 4y = 5**

Sol:

Given equations are 2x + 3y = 9 ———— (1)

3x + 4y = 5———– (2)

From (1)

3y = 9 – 2x

∴ the solution is x = – 21 and y = 17

**3. Solve the following pair of equations by using substitution method**

** 3x – 5y = – 1 and x – y = – 1**

Sol:

Given equations are 3x – 5y = – 1 ———— (1)

x – y = – 1———– (2)

From (2)

y = x + 1

substitute y value in (1)

⟹ 3x – 5(x + 1) = – 1

3x – 5x – 5= – 1

– 2x – 5= – 1

– 2x = – 1 + 5

– 2x = 4 ⟹ x = – 2

now y = x + 1

⟹ y = – 2 + 1 = – 1

∴ the solution is x = – 2 and y = – 1

**4. Solve the following pair of equations by using substitution method**

** x + = 6 and 3x – = 5**

**Sol:**

**5. Solve the following pair of equations by using substitution method**

** 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3**

**Sol:**

In this method first we eliminate one of the two variables by equating its coefficients. This give a single equation which can be solved to get the value of the other variable

Steps for Elimination Method:

- Write the both the equations in the form of ax + by = c
- Multiply both the equations by suitable non-zero real number to make the coefficient of one variable (x or y) equal
- If the variable to be eliminated has the same sign in both the equations, subtract one equation from another to get an equation in one variable.
- Solve the equation for the one variable
- Substitute the value of this variable in any one of the given equations and find the value of the eliminated variable.

**1. Solve the following pair of equations by using Elimination method**

** 2x – y = 5 and 3x + 2y = 11**

**2. Solve the following pair of equations by using Elimination method**

** 8x + 5y = 9 and 3x + 2y = 4**

**3. Solve the following pair of equations by using Elimination method**

** 3x + 4y = 25 and 5x – 6y = –9**

**4. Solve the following pair of equations by using Elimination method**

** 2x + y = 5 and 3x – 2y = 4**

**5. Solve the following pair of equations by using Elimination method**

** 3x + 2y = 11 and 2x + 4y =4**

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Education contributes to the development of society.

www.basicsinmaths.com website has been given material for mathematics Polytechnic Students.

Polytechnic Engineering Mathematics – Sem -1 Solutions solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.

Transformations PDF File is designed by the ‘Basics In Maths” Team. This Pdf File very useful for students who are prepared for polytechnic examinations.

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PART – 2

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Inverse trigonometric functions solutions PDF File is designed by the ‘Basics In Maths” Team. This Pdf File very useful for students who are prepared for polytechnic examinations.

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Solutions of simultaneous equations PDF File is designed by the ‘Basics In Maths” Team. This Pdf File very useful for students who are prepared for polytechnic examinations.

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Solutions of triangles PDF File is designed by the ‘Basics In Maths” Team. This Pdf File very useful for students who are prepared for polytechnic examinations.

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Education contributes to the development of society.

www.basicsinmaths.com website has been given material for math Polytechnic Students.

Engineering Maths Sem – 2 Concept solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.

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Education contributes to the development of society.

www.basicsinmaths.com website has been given material for mathematics Polytechnic Students.

Polytechnic Engineering Mathematics – Sem -2 Solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.

Tangents and Normals solutions PDF File is designed by the ‘Basics In Maths” Team. This Pdf File very useful for students who are prepared for polytechnic examinations.

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The Maxima and Minim solutions PDF File is designed by the ‘Basics In Maths” Team. This Pdf File is very useful for students who are prepared for polytechnic examinations.

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These Practice papers to do help the intermediate First-year Maths students.

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Inter Maths – 1A & 1B PDF Files are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.

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The English Language is important to communicate and interact with other people around us. It keeps us in contact with other people.

An example of the importance of a language is the ‘English language’ because it is the international language and has become the most important language to people in many parts of the world.

British brought with them their language English to India.

ఇంగ్లీష్ భాష:

మన చుట్టూ ఉన్న ఇతర వ్యక్తులతో సందేశించాదానికికి మరియు మన చుట్టూ ఉన్న ఇతర వ్యక్తులతో సంభాషించడానికి భాష ముఖ్యం. ఇది మనల్ని ఇతర వ్యక్తులతో సంప్రదించుటకు దోహదపడుతుంది.

ఒక భాష యొక్క ప్రాముఖ్యతకు ఒక ఉదాహరణ ‘ఆంగ్ల భాష’ ఎందుకంటే ఇది అంతర్జాతీయ భాష మరియు ప్రపంచంలోని అనేక ప్రాంతాల ప్రజలకు అత్యంత ముఖ్యమైన భాషగా మారింది.

బ్రిటిష్ వారు తమ భాష ఇంగ్లిష్ ను భారతదేశానికి తీసుకువచ్చారు.

Grammar is the way we arrange words to make proper sentences. Grammar rules about how to speak and write in a language. English grammar is the grammar of the English language. English grammar started out based on Old English,

ఇంగ్లిష్ వ్యాకరణం:

వ్యాకరణం అనేది సరైన వాక్యాలు చేయడానికి పదాలను ఏర్పాటు చేసే విధానం. వ్యాకరణం అనగా ఒక భాషలో ఎలా మాట్లాడాలి మరియు ఎలా రాయాలి అనే నియమాలు. ఆంగ్ల వ్యాకరణం ఆంగ్ల భాష యొక్క వ్యాకరణం. ఓల్డ్ ఇంగ్లిష్ ఆధారంగా ఇంగ్లిష్ గ్రామర్ ప్రారంభమైంది,

There are 26 letters in the English Language. Those are called as ‘Alphabet’

There are two parts to Alphabet.

- Vowels (a, e, i, o, u) [ 5 letters]
- Consonants (Remaining 21 letters)

Without vowel (sound or structure) we cannot create even a single word in English.

A noun is a naming word. The noun means the name of the person, things, places, or animals

(నామవాచకం ఒక వ్యక్తి యొక్క, ఒక వస్తువు యొక్క లేదా జంతువు యొక్క పేరును తెలుపుతుంది)

Ex: __Ramu__ goes to __college__ by __car __

__Seetha__ went to __school__ by __bus __ **→ underlined words are nouns**

**Proper Noun:** A proper noun denotes one particular person, place, or thing.

(Proper Noun, ఒక ప్రత్యేక వ్యక్తి, వస్తువు లేదా జంతువు యొక్క పేరును తెలుపుతుంది)

Ex: Raju, Hyderabad, The Ganga, etc.

**Common Noun: **A common noun is a name given commonly to a person, place or thing.

(CommonNoun, ఒకే జాతికి చెందిన వ్యక్తి, వస్తువు లేదా జంతువు యొక్క పేరును తెలుపుతుంది)

**Ex:** boy, girl, animal, river, city, etc.

**Collective noun:** ** **A collective noun denotes a group or collection of persons or things taken as one.

(Collective Noun, వ్యక్తుల, వస్తువుల లేదా జంతువుల యొక్క గుంపును తెలుపుతుంది)

Ex: herd, army, committee, flock, etc.,

**Material noun:** A Material noun denote the name of a particular kind of metal, liquid, or substance.

(Material Noun, ఒక నిర్దిష్ట రకం లోహం, ద్రవం లేదా పదార్థం యొక్క పేరును తెలియజేస్తుంది)

Ex: salt, sand, gold, rice, paddy, etc.,

**Concrete Noun: ** A Concrete noun denotes something that can be tasted, something that can be touched or seen, something that exists physically.

(కాంక్రీట్ నామవాచకం దేనినైనా రుచి చూడవచ్చు, ఏదైనా తాకవచ్చు లేదా చూడవచ్చు, భౌతికంగా ఉన్నదాన్ని సూచిస్తుంది.)

Ex: Pencil, boy, girl, gold, silver, rice, etc.,

**Note: **Proper nouns and Material nouns are Concrete nouns.

**Abstract Noun: **An Abstract noun denotes something maybe an idea or emotion.

Ex: born, sad, joy, bravery, freedom, etc.,

A pronoun is a word that is used instead of a noun.

సర్వనామం ను నామవాచకానికి బదులుగా వాడుతాము.

Ex: Ramu went to the Ground, __he__ played cricket.

పై వాక్యం లో రాము కు బదులుగా he వాడబడినది.

**Personal Pronouns:** Personal pronoun refers to a particular person or thing. (దీనిని వ్యక్తి పేరు కి బదులుగా ఉపయోగిస్తారు)

These are three types

ఇవి మూడు రకాలు

I person: Talks about himself (తన గురించి చెప్పేది. ఉదా : నేను, నాకు, మేము , మాకు, మొ ||)

Ex: I – we – my – us etc.,

II person: what it says about others (ఎదుటి వారి గురించి చెప్పేది. ఉదా : నీవు , మీరు , మీకు మొ ||)

Ex: you, yours

III Person: Talks about the third person between the discussion of two people (ఇద్దరి వ్యక్తుల సంభాషణ మధ్య మూడో వ్యక్తి గురించి చెప్పే ది. ఉదా : అతను , ఆమె , అతనికి , ఆమెకి , వారికి మొ ||)

Ex: he, she, it, they. Etc.,

Reflexive Pronouns are used when the subject and the object of a sentence are the same. They can act as either objects or indirect objects. (ఒక వ్యక్తి చేసిన పని ఫలితాన్ని తానే పొందినప్పుడు వీటిని వాడుతారు)** **

**Ex: **myself, himself, themself, yourselves

A relative pronoun introduces a clause. It refers to some noun going before and also joins two sentences together. (రెండు వాక్యములను కలుపుటకు వాడుతాము లేదా ఒక వాక్యములో అంతకుముందే చెప్పబడిన nouns ను refer చేస్తాయి)

**Ex: **who ……. Persons **కు**

** **Which ……. Places **కు**

** **That …… Things కు **వాడుతారు **

**Demonstrative Pronoun: **Demonstrative pronouns always identify nouns, whether those nouns are named specifically or not **(**ఇది, దేనినైనా లేక వేనినైన ఎత్తి చూపడానికి ఉపయోగపడుతుంది)

Ex: this, that, those, these, etc.,

**Distributive Pronoun: **Distributive pronouns refer to persons or things one at a time. (ఒకే సమయం లో ఎందరికో చెందేవి)

Ex; each, either, neither, etc.,

**Indefinite Pronouns:** Indefinite pronouns refer to people or things without saying exactly who or what they are. (ఫలానా వ్యక్తీ, ఫలానా వస్తువు గురించి కాకుండా ఎవరో ఒక వ్యక్తి, ఎదో ఒక వస్తువు గురించి Indefinite Pronouns తెలియజేస్తాయి)

Ex: somebody, none, all, nobody, etc.,

**Interrogative Pronouns: **These are used to ask questions (ప్రశ్నలు అడగడానికి వాడుతాము)

Ex: What, who, why

**Subject (కర్త ): **Subject means noun or pronoun or noun and pronoun.

An adjective is used with a noun to add something to its meaning (ఒక విశేషణం నామవాచకంతో దాని అర్థానికి ఏదైనా జోడించడానికి ఉపయోగించబడుతుంది)

Ex: large, big, small, honest, wise, etc.,

**Qualitative Adjective:** It indicates the characteristic of a person or an object (ఇది ఒక వ్యక్తి లేదా ఒక వస్తువు యొక్క లక్షణాన్ని తెలుపుతుంది)

Ex: honest, wise, small, big, etc.,

**Quantitative adjective: **It shows how much of a thing is (ఇవి ఎంత అనే అర్థంలో వాడుతాము)

Ex: some, much, little, enough, etc.,

**Numeral Adjectives: **It shows how many things are meant (సంఖ్యాత్మకమైనవి. ఎన్ని అనే పదానికి సమాధానంగా వచ్చేవి)

Ex: few, many, most, five, three, etc.,

**Demonstrative Adjectives:**

These, that, those, this వంటి నామవాచకం తో కలిపి వస్తే వాటిని Demonstrative Adjectives అంటారు.

Ex: This boy is tall

That girt is clever

Verb (క్రియ) లేకుండా ఇంగ్లీష్ లో వాక్యము లు ఉండవు. ఇంగ్లీష్ వాక్యానికి ‘క్రియ’ ప్రాణం వంటిది.

A verb is a word that tells something about an action or state.

Verb (క్రియ) లేకుండా ఇంగ్లీష్ లో వాక్యము లు ఉండవు. ఇంగ్లీష్ వాక్యానికి ‘క్రియ’ ప్రాణం వంటిది.

A verb is a word that tells something about an action or state. (పనులను తెలియజేయు పదాలను verbs అంటారు)

Ex: I __go__ to school

I __am__ a student

He __plays__ Cricket

We __sit__ at the table

There are two kinds of verbs:

- Main verb
- Helping verb

A verb that has an individual meaning is called the Main verb (వ్యక్తిగత అర్థాన్ని కలిగి ఉన్న క్రియను ప్రధాన క్రియ అని అంటారు)

Ex: go, come, take, sing, play, etc.,

**Helping verbs:**

A verb that does not have any individual meaning is called Helping verb (వ్యక్తిగత అర్థం లేని క్రియను Helping verb అని అంటారు)

Helping verb is mainly used to identify the tense. (ప్రధానంగా టెన్స్ గుర్తించడానికి Helping verb ఉపయోగించబడుతుంది)

Ex: do, does, Is, am, are, have, has, had, will, shall, etc.,

**Transitive Verb: **A transitive verb is a verb that denotes an action that passes over from the subject to an object. (Object ను కలిగి యుండే verb ను transitive verb అంటారు)

Ex: He writes a letter

Raju sings a song

**In Transitive Verb: ** An intransitive verb is a verb that denotes an action that does not pass over to an object. (Object లేని verb ను Intransitive verb అంటారు)

Ex: the boy plays

The Bird sings

An Adverb is a word that modifies the meaning of a verb, an adjective, or another adverb. (ఒక వాక్య్తం లోని ఒక verb, adjective లేదా adverb గురించి తెలిపేది)

Ex: quickly, very, quietly, clearly etc.,

A preposition is a word that is placed before a noun or pronoun to show the relation between a person, place, or thing. (ఒక నామవాచకముకు లేదా సర్వనామమునకు ముందున్చబడి వాక్యంలోని ఇతర పదం లేక పదాలతో అ నామవాచకం లేదా సర్వనామం యొక్క సంబందాన్ని తెలిపే పదం)

**Simple Prepositions: ** at, by, in, for, off, of, up, to, with, etc., are Simple Prepositions

**Compound Prepositions: **about, across, along, among, behind, before, below, besides, inside, within, without, etc., are called Compound Prepositions.

**Phrase Prepositions: **according to, Infront of, in favor of, because of, with regard to, etc., are known as Phrase Prepositions.

Conjunction combines sentences or words together. (కొన్ని వాక్యాలను లేదా పదాలను కలిపే పదం)

Ex: and, or, but, so, if, as, since, when, etc.,

An Interjection is a word that expresses some sudden feelings or emotions.

(మానసిక భావాలను లేక ఉద్రేకాలను తెల్పుటకు వాడే మాటలను Interjections అని అంటారు).

Note: Interjections తరువాత ఆశ్చర్యార్ధకము (!) అనే గుర్తు ఉంచి దాని తరువాత వచ్చేమాట మొదటి అక్షరము Capital letters తో ప్రారంభించవలెను.

Ex: Hello! What are you doing here?

Alas! He injured

Hurrah! I won the game

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**Logarithm:** For ant two positive real numbers a, b, and a ≠ 1. If the real number x such then a^{x} = b, then x is called logarithm of b to the base a. it is denoted by

Let a be a positive real number and a ≠ 1. The function f: (o, ∞) → R Defined by f(x) =

If f(x) and g(x) are two polynomials, g(x) ≠ 0, then is called rational fraction.

Ex:

_{ }etc. are rational fractions.

A rational fraction is said to be a Proper fraction if the degree of g(x) is greater than the degree of f(x).

Ex:

_{ }etc. are the proper fractions.

A rational fraction is said to be an Improper fraction if the degree of g(x) is less than the degree of f(x).

Ex:

_{ }etc. are the Improper fractions.

Expressing rational fractions as the sum of two or more simpler fractions is called resolving a given fraction into a partial fraction.

∎ If R(x) = is proper fraction, then

**Case(i)**: – For every factor of g(x) of the form (ax + b)^{ n}, there will be a sum of n partial fractions of the form:

**Case(ii): – **For every factor of g(x) of the form (ax^{2} + bx + c)^{ n}, there will be a sum of n partial fractions of the form:

∎ If R(x) = is improper fraction, then

**Case (i): – **If degree f(x) = degree of g(x), where k is the quotient of the highest degree term of f(x) and g(x).

**Case (ii): – **If f(x) > g(x)

**Matrix**: A set of numbers arranged in the form of a rectangular array having rows and columns is called Matrix.

•Matrices are generally enclosed by brackets like

•Matrices are denoted by capital letters A, B, C, and so on

•Elements in a matrix are real or complex numbers; real or complex real-valued functions.

**Oder of Matrix:** A matrix having ‘m’ rows and ‘n’ columns is said to be of order m x n read as m by n.

Ex:

**Rectangular Matrix: **A matrix in which the no. of rows is not equal to the no. of columns is called a rectangular matrix._{ }

** Square Matrix:** A matrix in which the no. of rows is equal to no. of columns is called a square matrix.

**Principal diagonal (diagonal) Matrix:** If A = [a _{ij}] is a square matrix of order ‘n’ the elements a_{11}, a_{22}, a_{33}, ………. A_{n n} is said to constitute its principal diagonal.

**Trace Matrix:** The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A).

**Diagonal Matrix:** If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix.

**Scalar Matrix:** If each non-diagonal element of a square matrix is ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix.

**Identity Matrix or Unit Matrix:** If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called unit matrix

**Null Matrix or Zero Matrix:** If each element of a matrix is zero, then it is called a null matrix.

**Row matrix & column Matrix:** A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix.

A square matrix A = [a_{ij}] is said to be upper triangular if a_{ij} = 0 ∀ i > j

A square matrix A = [a_{ij}] is said to be lower triangular matrix a_{ij} = 0 ∀ i < j

matrices A and B are said to be equal if A and B are of the same order and the corresponding elements of A and B are equal.

If A and B are two matrices of the same order, then the matrix obtained by adding the corresponding elements of A and B is called the sum of A and B. It is denoted by A + B.

If A and B are two matrices of the same order, then the matrix obtained by subtracting the corresponding elements of A and B is called the difference from A to B.

Let A = [a_{ik}]_{mxn} and B = [b_{kj}]_{nxp} be two matrices, then the matrix C = [c_{ij}]_{mxp} where

**Note:** Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix.

A _{m x n } . B_{p x q} = AB _{mx q}; n = p

**Transpose of Matrix:** If A = [a_{ij}] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by A^{I} or A^{T}.

**Note:** (i) (A^{I})^{I} = A (ii) (k A^{I}) = k . A^{I} (iii) (A + B )^{T} = A^{T} + B^{T} (iv) (AB)^{T} = B^{T}A^{T}

**Symmetric Matrix:** A square matrix A is said to be symmetric if A^{T }=A

If A is a symmetric matrix, then A + A^{T} is symmetric.

**Skew-Symmetric Matrix:** A square matrix A is said to be skew-symmetric if A^{T }= -A

If A is a skew-symmetric matrix, then A – A^{T} is skew-symmetric.

** Minor of an element:** Consider a square matrix

the minor element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present.

Ex: – minor of a_{3} is = b_{1}c_{2} – b_{2}c_{1}

Minor of b_{2 }is = a_{1}c_{3} – a_{3}c_{1}

**Cofactor of an element**: The cofactor of an element in i ^{th} row and j ^{th} column of A_{3×3 }matrix is defined as its minor multiplied by (- 1) ^{i+j}.

If each element of a row (column) f a square matrix is zero, then the determinant of that matrix is zero.

If A is a square matrix of order 3 and k is scalar then.

If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero.

If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants.

If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix.

For any square matrix A Det(A) = Det (A^{I}).

Det (AB) = Det(A). Det(B).

For any positive integer n Det (A^{n}) = (DetA)^{n}.

A Square matrix is said to be singular if its determinant is is zero, otherwise it is said to be non-singular matrix.

∴ A is singular matrix

Det(B) = 4 + 4 = 8≠ 0

∴ B is non-singular

** ****Adjoint of a matrix:** The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A.

Let A = and cofactor matrix of A =

** ****Invertible matrix:** Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is a unit matrix of the same order as A and B.

The algebraic sum of two or more angles is called a ‘compound angle’. Thus, angles A + B, A – B, A + B + C etc., are Compound Angles

For any two real numbers A and B

**⋇ **sin (A + B) = sin A cos B + cos A Cos B

**⋇ **sin (A − B) = sin A cos B − cos A Cos B

**⋇ **cos (A + B) = cos A cos B − sin A sin B

**⋇ **cos (A − B) = cos A cos B + sin A sin B

**⋇ **sin (A + B + C) = ∑sin A cos B cos C − sin A sin B sin C ** **

**⋇ **cos (A + B + C) = cos A cos B cos C− ∑cos A sin B sin C** **

⋇ sin (A + B) sin (A – B) = sin^{2 }A – sin^{2} B = cos^{2 }B – cos^{2} A

⋇ cos (A + B) cos (A – B) = cos^{2 }A – sin^{2} B = cos^{2 }B – sin^{2} A

If A is an angle, then its integral multiples 2A, 3A, 4A, … are called ‘multiple angles ‘of A and the multiple of A by fraction like are called ‘submultiple angles.

∎ If is not an add multiple of

In ∆ABC,

Lengths AB = c; BC = a; AC =b

The area of the triangle is denoted by ∆.

Perimeter of the triangle = 2s = a + b + c

A = ∠CAB; B = ∠ABC; C = ∠BCA.

R is circumradius.

In ∆ABC,

⟹ a = 2R sin A; b = 2R sin B; c = 2R sin C

Where R is the circumradius and a, b, c, are lengths of the sides of ∆ABC.

In ∆ABC,

In ∆ABC,

a = b cos C + c cos B

b = a cos C + c cos A

c = a cos B + b cos A

In ∆ABC, a, b, and c are sides

**⨂ **The function f: R→R defined by f(x) = ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by Sinh x.

Similarly,

**⨂ **cosh^{2}x – sinh^{2} x = 1

cosh^{2}x = 1 + sinh^{2} x

sinh^{2} x = cosh^{2} x – 1

**⨂ **sech^{2} x = 1 – tanh^{2} x

tanh^{2} x = 1 – sesh^{2} x

**⨂ **cosech^{2} x = coth^{2} x – 1

coth^{2} x = 1 + coth^{2} x

**⨂ **Sinh (x + y) = Sinh x Cosh y + Cosh x Sinh y

**⨂ **Sinh (x − y) = Sinh x Cosh y − Cosh x Sinh y

**⨂ **Cosh (x + y) = Cosh x Cosh y + Sinh x Sinh y** **

**⨂ **Cosh (x − y) = Cosh x Cosh y − Sinh x Sinh y** **

**⨂ **sinh 2x = 2 sinh x cosh 2x** = **

**⨂ **cosh 2x = cosh^{2}x + sinh^{2} x = 2 cosh^{2}x – 1 = 1 + 2 sinh^{2}x =

**Inverse hyperbolic functions:**

The equation x^{2} + 1 = 0 has no roots in real number system.

∴ scientists imagined a number ‘i’ such that i^{2} = − 1.

if x, y are any two real numbers then the general form of the complex number is

z = x + i y; where x real part and y is the imaginary part.

3 + 4i, 2 – 5i, – 3 + 2i are the examples for Complex numbers.

- z = x +i y can be written as (x, y).
- If z
_{1}= x_{1}+ i y_{1}, z_{2}= x_{2}+ i y_{2}, then - z
_{1 + }z_{2}= (x_{1}+ x_{2}, y_{1}+ y_{2}) = (x_{1}+ x_{2}) + i (y_{1}+ y_{2}) - z
_{1 − }z_{2}= (x_{1}− x_{2}, y_{1}− y_{2}) = (x_{1}− x_{2}) + i (y_{1}− y_{2}) - z
_{1∙ }z_{2}= (x_{1}x_{2}−y_{1}y_{2}, x_{1}y_{2 }+ x_{2}y_{1}) = (x_{1}x_{2}−y_{1}y_{2}) + i (x_{1}y_{2}+x_{2}y_{1}) - z
_{1/ }z_{2 = (}x_{1}x_{2}+ y_{1}y_{2}/x_{2}^{2}+y_{2}^{2}, x_{2}y_{1 }– x_{1}y_{2}/ x_{2}^{2}+y_{2}^{2})

= (x_{1}x_{2} + y_{1} y_{2}/x_{2}^{2} +y_{2}^{2}) + i (x_{2} y_{1 }– x_{1}y_{2}/ x_{2}^{2} +y_{2}^{2})

** The multiplicative** inverse of the complex number z is 1/z.

z = x + i y then 1/z = x – i y/ x^{2} + y^{2}

The complex numbers x + i y, x – i y are called conjugate complex numbers.

The sum and product of two conjugate complex numbers are real.

If z_{1}, z_{2} are two complex numbers then

**Modulus: –** If z = x + i y, then the non-negative real number is called modulus of z and it is denoted by or ‘r’.

**Amplitude: – **The complex number z = x + i y represented by the point P (x, y) on the XOY plane. ∠XOP = θ is called amplitude of z or argument of z.

x = r cos θ, y = r sin θ

x^{2} + y^{2} = r^{2} cos^{2}θ + r^{2} sin^{2}θ = r^{2} (cos^{2}θ + sin^{2}θ) = r^{2}(1)

⇒ x^{2} + y^{2} = r^{2}

• Arg (z) = tan^{−1}(y/x)

• Arg (z_{1}.z_{2}) = Arg (z_{1}) + Arg (z_{2}) + nπ for some n ∈ { −1, 0, 1}

• Arg(z_{1}/z_{2}) = Arg (z_{1}) − Arg (z_{2}) + nπ for some n ∈ { −1, 0, 1}

Note:

∎ e^{iθ} = cos θ + i sin θ

∎ e^{−iθ} = cos θ − i sin θ

For any integer n and real number θ, (cos θ + i sin θ)^{ n} = cos n θ + i sin n θ.

**→** cos α + i sin α can be written as cis α

**→** cis α.cis β= cis (α + β)

**→ **1/cisα = cis(-α)

**→ **cisα/cisβ = cis (α – β)

**⟹** (cos θ + i sin θ)^{ -n} = cos n θ – i sin n θ

**⟹** (cos θ + i sin θ) (cos θ – i sin θ) = cos^{2}θ – i^{2} sin^{2}θ = cos^{2}θ + sin^{2}θ = 1.

**→ **cos θ + i sin θ = 1/ cos θ – i sin θ and cos θ – i sin θ = 1/ cos θ + i sin θ

**⟹** (cos θ – i sin θ)^{ n} = (1/ (cos θ –+i sin θ))^{ n} = (cos θ + i sin θ)^{-n} = cos n θ – i sin n θ

**n ^{th} root of a complex number: **let n be a positive integer and z

**⟹** let z = r (cos θ + i sin θ) ≠ 0 and n be a positive integer. For k∈ {0, 1, 2, 3…, (n – 1)}

let Then a_{0}, a_{1}, a_{2, …, }a_{n-1} are all n distinct n^{th} roots of z and any n^{th} root of z is coincided with one of them.

**n ^{th} root of unity: **Let n be a positive integer greater than 1 and

**Note:**

- The sum of the n
^{th}roots of unity is zero. - The product of n
^{th}roots of unity is (– 1)^{ n – 1}. - The n
^{th}roots of unity 1, ω, ω^{2}, …, ω^{n-1}are in geometric progression with common ratio ω.

x^{3} – 1 = 0 ⇒ x^{3} = 1

x =1^{1/3}

ω^{2} +ω + 1 = 0 and ω^{3} = 1

For A, B∈ R

⋇ sin (A + B) + sin (A – B) = 2sin A cos B

⋇ sin (A + B) −sin (A – B) = 2cos A sin B

⋇ cos (A + B) + cos (A – B) = 2 cos A cos B

⋇ cos (A + B) − cos (A – B) = − 2sin A sin B

For any two real numbers C and D

If A + B + C = π or 180^{0}, then

⋇ sin (A + B) = sin C; sin (B + C) = sin A; sin (A + C) = sin B

⋇ cos (A + B) = − cos C; cos (B + C) = −cos A; cos (A + C) = − cos B

⋇ sin = cos; sin = cos ; sin = cos

⋇ cos = sin; cos = sin; cos = sin

⋇ sin (A + B) = cos C; sin (B + C) = cos A; sin (A + C) = cos B

⋇ cos (A + B) = sin C; cos (B + C) = sin A; cos (A + C) = sin B

If A, B are two sets and f: A→ B is a bijection, then f^{-1 }is existing and f^{-1}: B → A is an inverse function.

Let a system of simultaneous equations be

a_{1 }x + b_{1} y + c_{1}z = d_{1}

a_{2 }x + b_{2} y + c_{2}z = d_{2}

a_{3 }x + b_{3} y + c_{3}z = d_{3}

The matrix form of the above equations is

Therefore, the matrix equation is AX = B

If Det A ≠ 0, A^{-1 }is exists

X = A^{-1 }B

By using above Condition, we get the values of x, y and z

This Method is called as Matrix Inversion Method

Let system of simultaneous equations be

a_{1 }x + b_{1} y + c_{1}z = d_{1}

a_{2 }x + b_{2} y + c_{2}z = d_{2}

a_{3 }x + b_{3} y + c_{3}z = d_{3}

∆_{1 }is obtained by replacing the coefficients of x (1^{st} column elements of ∆) by constant values

∆_{2 }is obtained by replacing the coefficients of y (2^{nd} column elements of ∆) by constant values

∆_{3 }is obtained by replacing the coefficients of z (3^{rd} column elements of ∆) by constant values

This method is called Cramer’s Method

Let a system of simultaneous equations be

a_{1 }x + b_{1} y + c_{1}z = d_{1}

a_{2 }x + b_{2} y + c_{2}z = d_{2}

a_{3 }x + b_{3} y + c_{3}z = d_{3}

**Augmented matrix:** The coefficient matrix (A) augmented with the constant column matrix (B) is called the augmented matrix. It is denoted by [AD].

This Matrix is reduced to the standard form ofby using row operations

- Interchanging any two rows
- Multiplying the elements of any two elements by a constant.
- Adding to the elements of one row with the corresponding elements of another row multiplied by a constant.

∴ The solution of a given system of simultaneous equations is x = α, y = β, and z = γ.

- Take the coefficient of x as the unity as a first equation.
- If 1 is there in the first-row first column, then make the remaining two elements in the first column zero.
- After that, if one element in R
_{2}or R_{3}is 1, then make the remaining two elements in that column C_{2}or C_{3}as zeroes. - If any row contains two elements as zeros and only non-zero divide that row elements with the non-zero element to get unity and make the remaining two elements in that column as zeros.

PDF FILE: Mathematics Notes 4 Polytechnic SEM – I

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Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

**Errors and Approximations**

Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions

**Sol: **

Given y = f(x) = x^{2} + x at x = 10, ∆x = 0.1

∆y = f (x + ∆x) – f (x)

= f (10 + 0.1) – f (10)

= f (10.1) – f (10)

= (10. 1)^{2} + 10.1 – (10^{2} + 10)

= 102.01 + 10.1 – 100 – 10

= 112.11 – 110

= 2.11

dy = f’ (x) ∆x

= (2x + 1) (0.1)

= [2(10) + 1] (0.1)

= 21 × 0.1

= 2.1

**Sol: **

Given y = cos x, x = 60^{0} and ∆x = 1^{0}

∆y = f (x + ∆x) – f (x)

= cos (60^{0} + 1^{0}) – cos 60^{0}

= cos (61^{0}) – 0.5

= 0.4848 – 0.5

= – 0.0152

dy = f’ (x) ∆x

= – sin x (1^{0})

= – sin 60^{0} × 0.0174

=– 0.8660 × 0.0174

= – 0.0150

**Sol: **

y = x^{2} + 3x + 6

∆y = f (x + ∆x) – f (x)

= f (10 + 0.01) – f (10)

= f (10.01) – f (10)

= (10.01)^{2} + 3 (10.01) + 6 – (10^{2} + 3 (10) + 6)

= 100. 2001 + 30.03 + 6 – 100 – 30 – 6

=130. 2301 – 130

= 0.2301

dy = f’ (x) ∆x

= (2x + 3 + 0) (0.01)

= (2× 10 + 3) (0.01)

= 23 × 0.01

= 0.23

**Sol: **

The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square.

**Sol:**

Let x be the side of the square and area be A

Area of the square A = x^{2}

** **x = 3 and ∆x = 0.01

∆A = 2x × ∆x

= 2(3) (0.01)

= 6 × 0.01

= 0.06

If an increase in the side of a square is 2% then find the approximate percentage of increase in its area.

**Sol: **

Let x be the side of the square and A be its area

A = x^{2}

∆A = 2x × ∆x

The approximate percentage error in area A

= 2 × 2 =4

From the following. Find the approximations

**Sol: **

Let f(x) = , where x = 1000 and ∆x =– 1

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= 10 – 0. 0033

= 9.9967

**Sol:**

**Sol:**

**(iv) **Sin 62^{0}

**Sol: **

Let f(x) = sin x, where x = 60^{0} and ∆x =2^{0}

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 60^{0} + cos x (2^{0})

= sin 60^{0}+ cos 60^{0} (0.0348)

= 0.8660 + 0.5 × 0.0348

= 0.8660 + 0.0174

=0.8834

The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere.

**Sol:**

Given r = 14 cm and ∆r =0.02cm

Surface area of sphere =A = 4π r^{2}

∆A = 8π r ∆r

= 8 ×3.14× 14 × 0.02

= 7.0336

]]>

Find f’ (x) for the following functions

**(i)** f(x) = (ax + b) (x > -b/a)

**Sol:**

Given f(x) = (ax + b)^{ n}

f’ (x) = n (ax + b)^{ n – 1 } (ax + b)

= n (ax + b) ^{n – 1 }a

= an (ax + b) ^{n –} 1

**(ii) ** f(x) = x^{2} 2^{x} log x

**Sol:**

Given f(x) = x^{2} 2^{x} log x

f’ (x) = ^{ }(x^{2}) 2^{x} log x + x^{2} (2^{x}) log x + x^{2} 2^{x} (log x).

= 2×2^{x} log x +x^{2} 2^{x} log a log x + x^{2} 2^{x} (1/x)

= x 2^{x}[log x^{2} + x log x log 2 + 1]

**Sol:**

f’ (x) = . log 7 ^{ }(x^{3} + 3x)

**(iv) **f(x) = log (sec x + tan x)

**Sol:**

Given, f(x) = log (sec x + tan x)

= sec x

Find the derivative of the following functions

**(i) **f(x) = e^{x} (x^{2} + 1)

**Sol:**

Given f(x) = e^{x} (x^{2} + 1)

f’ (x) = e^{x} (x^{2} + 1) + (x^{2} + 1) ^{ }(e^{x})

= e^{x} (2x + 0) + (x^{2} + 1) e^{x}

= e^{x} (x^{2} + 2x + 1)

= e^{x} (x + 1)^{2}

**(iii) **cos (log x + e^{x})

**(iv) **x = tan (e^{-y})

e^{-y} = tan^{-1} x

**(v) **cos [log (cot x)]

**(vi) **sin[tan^{-1}(e^{x})]

**(vii) **cos^{-1}(4x^{3} – 3x)

let y = cos^{-1}(4x^{3} – 3x)

put x = cos θ ⟹ θ = cos^{-1} x

y = cos^{-1}(4 cos^{ 3} θ – 3cos θ)

= cos^{-1}(cos 3θ)

= 3 θ

= 3 cos^{-1} x

Find f’ (x), If f(x) = (x^{3} + 6 x^{2} + 12x – 13)^{100}.

**Sol:**

Given f(x) = (x^{3} + 6 x^{2} + 12x – 13)^{100}

f’ (x) = 100(x^{3} + 6 x^{2} + 12x – 13)^{99} (x^{3} + 6 x^{2} + 12x – 13)

= 100(x^{3} + 6 x^{2} + 12x – 13)^{99} (3x^{2} + 12 x + 12 – 0)

=100(x^{3} + 6 x^{2} + 12x – 13)^{99} 3 (x^{2} + 4 x + 4)

= 300 (x + 2)^{2} (x^{3} + 6 x^{2} + 12x – 13)^{99}

If f(x) = 1 + x + x^{2} + x^{3} + …. + x^{100}, then find f’ (1).

**Sol:**

Given f(x) = 1 + x + x^{2} + x^{3} + …. + x^{100}

f’(x) = 0 + 1 + 2x + 3 x^{2} + … 100 x^{99}

f’(1) = 1 + 2 + 3 + … + 100

= 50 × 101

= 5050

From the following functions. Find their derivatives.

**Sol:**

**Sol:**

Given y = log (cosh 2x)

If x = a cos^{3} t, y = a sin^{3} t, find

**Sol:**

Given If x = a cos^{3} t, y = a sin^{3} t

Differentiate f(x) with respect to g(x) for the following.

derivative of f(x) with respect to g(x) =

put x = tan θ ⟹ θ = tan^{-1} x

**Sol: **

]]>

**Limits**

**Question 1**

**Sol:**

= 9

**Question 2**

**Sol:**

= a + a = 2a

**Question 3**

**Sol:**

**Question 4**

**Sol:**

we know that if x > 0

As x → 0+ ⟹ x > 0

As x → 0+ ⟹ x < 0

**Question 5**

If f (x) = , then find and . Does exist?

**Sol:**

**Question 6**

**Sol:**

As x → 2– ⟹ x < 2

x – 2 < 0

**Question 7**

**Sol: **

As x → 2+ ⟹ x > 2

As x → 2– ⟹ x < 2

**Question 8**

**Sol:**

**Question 9**

**Sol:**

**Question 10**

**Sol: **

**Question 11**

**Sol: **

**Question 12**

**Sol:**

**Question 13**

**Sol:**

let y = x – 1 ⟹ x = y + 1

then as x → 1, y → 0

**Question 14**

**Sol:**

**Question 15**

**Sol:**

**Question 16**

**Sol: **

As x → ∞, →0

= ∞ (3/4)

= ∞

**Question 17**

**Sol:**

We know that – 1 ≤ sin x ≤ 1

x^{2} – 1 ≤ x^{2} – sin x ≤ x^{2} + 1

**Question 18**

**Sol:**

**Question 19**

**Sol:**

**Question 20**

**Sol:**

**Question 21**

**Sol: **

**Question 22**

**Sol: **

**Question 23**

**Sol:**

]]>

**Hyperbolic Functions**

Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh^{3} x

**Sol:**

sinh (3x) = sinh (2x + x)

= sinh 2x cosh x + cosh 2x sinh x

= (2 sinh x cosh x) cosh x + (1 + 2 sinh^{2} x) sinh x

= 2sinh x cosh^{2} x + sinh x + 2 sinh^{3} x

= 2 sinh x (1 + sinh^{2} x) + sinh x + 2 sinh^{3} x

= 2 sinh x + 2 sinh^{3} x+ sinh x + 2 sinh^{3} x

= 3 sinh x + 4 sinh^{3} x

If cosh x = , find the values of (i) cosh 2x and (ii) sinh 2x

**Sol:**

Cosh 2x = 2 cosh^{2} x – 1

Sinh^{2} 2x = cosh^{2} 2x – 1

If cosh x = sec θ then prove that tanh^{2}= tan^{2}

**Sol: **

If sinh x = 5, then show that x =

**Sol: **

Given, sinh x = 5

⟹ x = sinh^{-1}5

**Sol: **

Given tanh^{-1}

For x, y ∈ R prove that sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)

**Sol: **

R.H.S = sinh (x) cosh (y) + cosh (x) sinh (y)

= sinh (x + y)

For any x∈ R, prove that cosh^{4} x – sinh^{4} x = cosh 2x

**Sol:**

cosh^{4} x – sinh^{4} x = (cosh^{2} x)^{2} – (sinh^{2} x)^{2}

= (cosh^{2} x + sinh^{2} x) (cosh^{2} x – sinh^{2} x)

= 1. cosh 2x

= cosh 2x

**Sol: **

**=** cosh x + sinh x

If sin hx = ¾ find cosh 2x and sinh 2x.

**Sol: **

Given sin hx = ¾

We know that cosh^{2} x = 1 + sinh^{2} x

= 1 + (3/4)^{2}

= 1 + 9/16

= 25/16

cos hx = 5/4

cosh 2x = 2cosh^{2} x – 1

= 2(25/16) – 1

= 25/8 – 1

= 17/8

Sinh 2x = 2 sinh x cosh x

= 2 (3/4) (5/4)

= 15/8

Prove that (cosh x – sinh x)^{ n} = cosh nx – sinh nx

**Sol: **

∴ (cosh x – sinh x)^{ n} = cosh nx – sinh nx

]]>

**Trigonometric Ratios Up to Transformations**

Find the value of sin^{2}(π/10) + sin^{2}(4π/10) + sin^{2}(6π/10) + sin^{2}(9π/10)

**Sol:**

** **sin^{2}(π/10) + sin^{2}(4π/10) + sin^{2}(6π/10) + sin^{2}(9π/10)

= sin^{2}(π/10) + sin^{2}(π/2 – π/10) + sin^{2}(π/2+ π/10) + sin^{2}(π – π/10)

= sin^{2}(π/10) + cos^{2}(π/10) + cos^{2}(π/10) + sin^{2}(π/10)

= 1 + 1 = 2

If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ

**Sol: **

Given sin θ = 4/5 and θ not in the first quadrant

⇒ θ in the second quadrant

⇒ cos θ < 0

cos^{2}θ = 1 – sin^{2} θ

=1 – (4/5)^{2}

= 1 – 16/25

∴cos θ = – 3/5 (∵cos θ < 0)

If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ

**Sol:**

Given, 3sin θ + 4 cos θ = 5

let 4 sin θ – 3cos θ = x

(3sin θ + 4 cos θ )^{2} + (4 sin θ – 3cos θ)^{2} = 5^{2} + x^{2}

9 sin^{2} θ + 16 cos^{2} θ + 12 sin θ cos θ + 16 sin^{2} θ + 9 cos^{2} θ – 12sin θ cis θ = 25 + x^{2}

25 sin^{2} θ + 25 cos^{2} θ = 25 + x^{2}

25 = 25 + x^{2}

⇒ x^{2} = 0

x = 0

∴ 4 sin θ – 3cos θ = 0

If sec θ + tan θ =, find the value of sin θ and determine the quadrant in which θ lies

**Sol:**

Given, sec θ + tan θ = ———— (1)

We know that sec^{2} θ – tan^{2} θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

(1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) =

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) =

Since sec θ positive and tan θ is negative θ lies in the 4^{th} quadrant.

Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1

**Sol:**

cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)

= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)

= 1.1.1.1

=1

If cos θ + sin θ = cos θ, then prove that cos θ – sin θ = sin θ

**Sol:**

( + 1) sin θ = ( + 1) ( – 1) cos θ

Find the value of 2(sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ)

**Sol:**

2(sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ)

= 2[(sin^{2} θ)^{3} + (cos^{2} θ)^{3}] – 3[(sin^{2} θ)^{2} + (cos^{2})^{2}

= 2[(sin^{2} θ + cos^{2} θ)^{3} – 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)] – 3[(sin^{2} θ + cos^{2} θ)^{2} – 2 sin^{2} θ cos^{2} θ]

= 2[1 – 3 sin^{2} θ cos^{2} θ] – 3 [1 – 2 sin^{2} θ cos^{2} θ]

= 2 – 6 sin^{2} θ cos^{2} θ – 3 + 6 sin^{2} θ cos^{2} θ

= – 1

If tan 20^{0} = λ, then show that

**Sol:**

Given tan 20^{0} = λ

If sin α + cosec α = 2, find the value of sin^{n} α + cosec^{n} α, n∈ Z

**Sol:**

Given sin α + cosec α = 2

⇒ sin α + 1/ sin α = 2

sin^{2} α + 1= 2 sin α

sin^{2} α – 2 sin α + 1= 0

(sin α – 1 )^{2} = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

sin^{n} α + cosec^{n} α = (1)^{n} + (1)^{n} =1 + 1 =2

∴ sin^{n} α + cosec^{n} α = 2

Evaluate sin^{2} **+ cos**^{2} ** – tan**^{2}

**Sol:**

Find the value of sin 330^{0}. cos 120^{0} + cos 210^{0}. Sin 300^{0}

**Sol:**

** **sin 330^{0}. cos 120^{0} + cos 210^{0}. Sin 300^{0}

**=**sin (360^{0} – 30^{0}). cos (180^{0} – 60^{0}) + cos (180^{0} + 30^{0}). sin (360^{0} – 60^{0})

**= **(– sin 30^{0}). (– cos 60^{0}) + (– cos30^{0}). (– sin60^{0})

**= **sin 30^{0}. cos 60^{0} + cos30^{0}. Sin60^{0}

**= **sin (60^{0} + 30^{0}) = sin 90^{0}

**=**1

Prove that cos^{4} α + 2 cos^{2} α = (1 – sin^{4} α)

**Sol: **

= cos^{4} α + 2 cos^{2} α (1 – cos^{2} α)

= (cos^{2} α)^{2} + 2 (1 – sin^{2} α) (sin^{2} α)

= (1 – sin^{2} α)^{2} + 2 sin^{2} α – 2sin^{4} α

= 1 + sin^{4} α – 2 sin^{2} α + 2 sin^{2} α – 2sin^{4} α

= 1 – sin^{4} α

Eliminate θ from x = a cos^{3} θ and y = b sin^{3} θ

**Sol: **

Given x = a cos^{3} θ and y = b sin^{3} θ

cos^{3} θ = x/a and sin^{3} θ = y/b

cos θ = (x/a)^{1/3} and sin θ = (y/b)^{1/3}

we know that sin^{2} θ + cos^{2} θ = 1

⇒ [(y/b)^{1/3}]^{2} + [(x/a)^{1/3}]^{2} = 1

(x/a)^{2/3} + (y/b)^{2/3} = 1

Find the period of the following functions

**Sol:**

**(i)** f(x) = tan 5x

we know that period of tan kx =

period of given function is = LCM (8, 6) = 24

**(iv)** f(x) = tan (x + 4x + 9x +…. + n^{2}x)

f(x) = tan (x + 4x + 9x +…. + n^{2}x)

= tan (1 + 4 + 9 + … + n^{2}) x

we know that period of tan kx =

Prove that sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0} =

**Sol:**

We know that sin^{2} A – sin^{2}B = sin (A +B) sin (A – B)

⇒ sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0}

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

= sin 75^{0} sin 30^{0}

∴ sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0} =

Prove that tan 70^{0} – tan20^{0} = 2 tan 50^{0}

**Sol:**

50^{0} = 70^{0} – 20^{0}

Tan 50^{0 }= tan (70^{0} – 20^{0})

⇒ tan 70^{0} – tan 20^{0} = tan 50^{0} (1 + tan70^{0} tan 20^{0})

tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + tan70^{0} cot (90^{0} – 20^{0})]

tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + tan70^{0} cot 70^{0}]

tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + 1]

∴ tan 70^{0} – tan20^{0} = 2 tan 50^{0}

If sin α = , sin β = and α, β are acute, show that α + β =

**Sol: **

tan α = 1/3 tan β = ½

tan (α + β) = 1

**Sol:**

**Sol:**

(on dividing numerator and denominator by cos 9^{0})

= tan (45^{0} + 9^{0})

= tan 54^{0}

= tan (90^{0} – 36^{0})

= cot 36^{0}

Show that cos 42^{0} + cos 78^{0} + cos 162^{0} = 0

**Sol:**

cos 42^{0} + cos 78^{0} + cos 162^{0}

= cos (60^{0} – 18^{0}) + cos (60^{0} + 18^{0}) + cos (180^{0} – 18^{0})

=cos 60^{0} cos18^{0} + sin 60^{0} sin 18^{0} + cos 60^{0} cos 18^{0} – sin 60^{0} sin 18^{0} – cos 18^{0}

= 2 cos 60^{0} cos 18^{0} – cos 18^{0}

= 2 (1/2) cos 18^{0} – cos 18^{0}

= cos 18^{0} – cos 18^{0}

= 0

Express sin θ + cos θ as a single of an angle

**Sol:**

** **sin θ + cos θ = 2( sin θ + cos θ)

** **= 2(cos 30^{0} sin θ + sin 30^{0} cos θ)

= 2 sin (θ + 30^{0})

Find the maximum and minimum value of the following functions

**(i) **3 sin x –4 cos x

a= 3, b = –4 and c = 0

= 5

∴ minimum value = –5 and maximum value = 5

(**ii) **cos (x + ) + 2 sin (x + ) – 3

a= 1, b = 2 and c = – 3

∴ minimum value = –6 and maximum value = 0

** Question 23**

**Sol: **

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

∴ Range = [–20, 30]

Prove that sin^{2}α + cos^{2} (α + β) + 2 sin α sin β cos (α + β) is independent of α

**Sol:**

sin^{2}α + cos^{2} (α + β) + 2 sin α sin β cos (α + β)

= sin^{2}α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin^{2}α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin^{2}α + cos (α + β) [ cos α cos β + sin α sin β]

=sin^{2}α + cos (α + β) cos (α –β)

= sin^{2} α + cos^{2} α – sin^{2} β

=1 – sin^{2} β

= cos^{2} β

**Sol:**

= tan θ

For what values of x in the first quadrant is positive?

**Sol:**

⟹ 0 < 2x < π/2 (∵ x is in first quadrant)

⟹ 0 < x < π/4

If cos θ = and π < θ < 3π/2, find the value of tan θ/2.

**Sol:**

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

= – 2

If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.

**Sol:**

= 2 cot 2A

Evaluate 6 sin 20^{0} – 8sin^{3} 20^{0}

**Sol:**

6 sin 20^{0} – 8sin^{3} 20^{0} = 2 (3 sin 20^{0} – 4sin^{3} 20^{0})

= 2 sin 3(20^{0})

= 2 sin 60^{0}

Express cos^{6} A + sin^{6} A in terms of sin 2A.

**Sol:**

cos^{6} A + sin^{6} A

**= **(sin^{2} A)^{3} + (cos^{2} A)^{3}

= (sin^{2} A + cos^{2} A)^{3} – 3 sin^{2} A cos^{2} A (sin^{2} A + cos^{2} A)

= 1 – 3 sin^{2} A cos^{2} A

=1 – ¾ (4 sin^{2} A cos^{2} A)

= 1 – ¾ sin^{2}2 A

If 0 < θ < π/8, show that = 2 cos (θ/2)

**Sol:**

=2 cos (θ/2)

Find the extreme values of cos 2x + cos^{2}x

**Sol:**

cos 2x + cos^{2}x = 2cos^{2} x– 1 + cos^{2} x

=3cos^{2} x – 1

We know that – 1 ≤ cos x ≤ 1

⟹ 0 ≤ cos^{2} x ≤ 1

3×0 ≤ 3×cos^{2} x ≤ 3×1

0– 1 ≤3 cos^{2} x – 1≤ 3– 1

– 1≤3 cos^{2} x – 1≤ 2

Minimum value = – 1

Maximum value = 2

**Sol:**

**= **4

Prove that sin 78^{0} + cos 132^{0} =

**Sol:**

** **sin 78^{0} + cos 132^{0} = sin 78^{0} + cos (90^{0} + 42^{0})

= sin 78^{0} – sin 42^{0}

= 2 cos 60^{0} sin 18^{0}

Find the value of sin 34^{0} + cos 64^{0} – cos4^{0}

**Sol:**

sin 34^{0} + cos 64^{0} – cos4^{0}

= sin 34^{0} – 2sin 34^{0} sin 30^{0}

= sin 34^{0} – 2 sin 34^{0} (1/2)

=sin 34^{0} – sin 34^{0}

=0

Prove that 4(cos 66^{0} + sin 84^{0}) =

**Sol:**

4(cos 66^{0} + sin 84^{0})

=4(cos 66^{0} + sin (90^{0 }– 6^{0})

=4(cos 66^{0} + cos (6^{0})

=8 cos 36^{0} cos 30^{0}

Prove that (tan θ + cot θ)^{2} = sec^{2} θ + cosec^{2} θ = sec^{2} θ. cosec^{2} θ

**Sol:**

= sec θ. cosec θ

(tan θ + cot θ)^{2} = sec^{2} θ. cosec^{2} θ

= sec^{2} θ. cosec^{2} θ

]]>

**Question 1**

If **a** = 6**i** +2 **j** +3 **k**, **b** = 2**i** – 9 **j+** 6**k**, then find the angle between the vectors **a** and **b**

**Sol: **

Given vectors are **a** = 6**i** +2 **j** +3 **k**, **b** = 2**i** – 9 **j+** 6**k**

If θ is the angle between the vectors **a** and **b**, then cos θ =

**a **.**b = **(6**i** +2 **j** +3 **k**). (2**i** – 9 **j+** 6**k**) = 6(2) + 2 (– 9) + 3(6)

** = **12 – 18 + 18 = 12

= 7

= 11

**Question 2**

If **a** = **i** +2 **j** –3 **k**, **b** = 3**i** – **j+** 2**k**, then show that **a** + **b **and **a** – **b** are perpendicular to each other.

**Sol: **

Given vectors are **a** = **i** +2 **j** –3 **k**, **b** = 3**i** – **j+** 2**k**

**a** + **b **= (**i** +2 **j** –3 **k**) + (3**i** – **j+** 2**k**) = 4**i** + **j** – **k**

**a** – **b** = (**i** +2 **j** –3 **k**) – (3**i** – **j+** 2**k**) = –2**i** +3 **j** – 5**k**

(**a** + **b**). (**a** – **b**) = (4**i** + **j** – **k**). (–2**i** +3 **j** – 5**k**)

= – 8 + 3 + 5

= 0

∴ **a** + **b **and **a** – **b** is perpendicular to each other.

If **a** and **b **be non-zero, non-collinear vectors. If , then find the angle between **a** and **b**

**Sol: **

Squaring on both sides

(**a** + **b**) (**a** + **b**) = (**a – b**) (**a – b**)

**a**^{2} + 2 **a**. **b** + **b**^{2} = **a**^{2} – 2 **a**.**b** + **b**^{2}

** **⟹ 4 **a**.**b** = 0

**a**.**b** = 0

∴ the angle between **a** and **b ** is 90^{0}

If = 11, = 23 and = 30, then find the angle between the vectors** a** and **b** and also find

**Sol: **

(11)^{2} **– **2 ×11×23 cos θ + (23)^{2} = 900

121 **– **506 cos θ + 529 = 900

650 **– **506 cos θ = 900

= (11)^{2} **+ **2 ×11×23 cos θ + (23)^{2}

= 121 + 2 ×11×23 × + 529

= 400

If **a** = **i** – **j** – **k** and **b** = 2**i** – 3**j** + **k**, then find the projection vector of **b** on **a** and its magnitude.

**Sol: **

Given vectors are **a** = **i** – **j** – **k** and **b** = 2**i** – 3**j** + **k**

** **a.b = (**i** – **j** –** k**). (2**i** – 3**j** + **k**) = 2 + 3 – 1 = 4

** **The projection vector of **b** on **a = **

** The magnitude** of the projection vector = =

If the vectors λ **i** – 3**j** + 5**k** and 2λ **i** – λ **j** – **k **are perpendicular to each other, then find λ

**Sol:**

let **a **= λ **i** – 3**j** + 5**k**, **b** = 2λ **i** – λ **j** – **k**

Given, that **a **and **b **are perpendicular to each other

⟹ **a**.**b** = 0

(λ **i** – 3**j** + 5**k**). (2λ **i** – λ **j** – **k**) = 0

2 λ^{2} + 3 λ – 5 = 0

2 λ^{2} + 5 λ – 2 λ – 5 = 0

λ (2 λ + 5) – 1 (2 λ + 5) = 0

(2 λ + 5) ((λ – 1) = 0

λ = 1 or λ = -5/2

Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3**i** + **j** + 5**k**

**Sol:**

let P (x, y, z) be any point on the plane

⟹ **OP** = x**i** + y**j** + z**k**

**OA** = – **2i** +**j** +3**k**

**AP** = **OP** – **OA** = (x**i** + y**j** + z**k**) – (– **2i** +**j** +3**k**)

** AP **= (x + 2) **i **+ (y – 1) **j** + (z – 3) **k**

** AP** is perpendicular to the vector 3**i** + **j** + 5**k **

⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0

⟹ 3x + 6 + y – 1 + 5z – 15 = 0

∴ 3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane

Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4

**Sol:**

Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4

Vector equations of the above planes are: **r**. (2**i** – 3**j** – 6**k**) = 5 and **r**. (6**i** + 2**j** – 9**k**) = 4

⟹ **n _{1}** = 2

If θ is the angle between the planes **r**. **n _{1}** = d

**a** = 2**i** – **j** + **k**, **b** = **i** – 3**j** – 5**k**. Find the vector **c** such that **a**, **b**, and **c** form the sides of a triangle.

**Sol:**

Given **a** = 2**i** – **j** + **k**, **b** = **i** – 3**j** – 5**k**

** **If a, **b**, and **c** form the sides of a triangle, then **a** + **b** + **c** = 0

⟹ **a** + **b **= – c

⟹ **c** = – (**a** + **b**)

= – [(2**i** – **j** + **k)** +( **i** – 3**j** – 5**k**)]

= – (3**i** –4 **j** –4**k**)

∴ **c** = – 3**i** +4 **j** + 4**k**

Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).

**Sol:**

Let **a** = 3**i** – 2**j** +** k **and **n = **4**i **+ 7**j **– 4**k **

** **The equation of the plane passing through point A(**a**) and perpendicular to the vector **n **is (**r** – **a**). **n** = 0

⟹ [**r** – (3**i** – 2**j** +** k**)]. (4**i **+ 7**j **– 4**k**) = 0

⟹ **r**. (4**i **+ 7**j **– 4**k**)– [(3**i** – 2**j** +** k**). (4**i **+ 7**j **– 4**k**)] = 0

**r**. (4**i **+ 7**j **– 4**k**)– (12 – 14 – 4) = 0

**r**. (4**i **+ 7**j **– 4**k**)– 6 = 0

**r**. (4**i **+ 7**j **– 4**k**) = 6

Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4**i** – 3**j** + **k**

**Sol: **

The vector which is parallel to the XOY-plane is of the form x**i** + y**j **

The vector which is parallel to the XOY-plane and perpendicular to 4**i** – 3**j** + **k**

is 3**i** + 4**j **

∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4**i** – 3**j** + **k **=

If **a** + **b** + **c** = 0, = 3, = 5 and = 7, then find the angle between **a** and **b**

**Sol:**

Given, **a** + **b** + **c** = 0, = 3, = 5 and = 7

**a** + **b** = – **c**

3^{2} + 5^{2} + 2 cos θ = 7^{2}

9 + 25 + 2.3.5 cos θ = 49

34 + 30cos θ = 49

30cos θ = 49 – 34

30cos θ = 15

cos θ = 15/30 = 1/2

∴ θ = π/3

If **a** = 2**i** – 3**j** + 5**k**, **b** = – **i** + 4**j** + 2**k**, then find **a** × **b** and unit vector perpendicular to both **a** and **b**

**Sol:**

Given,** a** = 2**i** – 3**j** + 5**k**, **b** = – **i** + 4**j** + 2**k **

** = i **(–6 – 20) – **j** (4 + 5) + **k** (8 – 3)

** = **–26**i – **9**j **+ 5**k**

The unit vector perpendicular to both **a** and **b** =

If **a** = **i** + **j** + 2**k** and **b** = 3**i** + 5**j** – **k** are two sides of a triangle, then find its area.

**Sol:**

Given, **a** = **i** + 2**j** + 3**k** and **b** = 3**i** + 5**j** – **k**

If** a, b **are two sides of a triangle, then area of the triangle =

** = i **(–2 – 15) – **j** (–1 – 9) + **k** (5 – 6)

** = **–17**i + **10** j **– **k**

Find the area of the parallelogram for which the vectors **a** = 2**i** – 3**j **and **b** = 3**i** – **k **are adjacent sides.

**Sol: **

Given, **a** = 2**i** – 3**j **and **b** = 3**i** – **k **are adjacent sides of a parallelogram

The area of the parallelogram whose vectors **a** , **b** are adjacent sides =

**= i **(3 – 0) – **j** (–2 – 0) + **k** (0 + 9)

** ** **=**3** i +**2 **j** +9 **k**

∴ The area of the parallelogram =

Let **a**, **b** be two non-collinear unit vectors. If α = **a** – (**a** . **b**) **b** and β = **a** × **b**, then show that

**Sol:**

= 1 – cos^{2} θ

= sin^{2} θ

= 1 + cos^{2} θ – 2cos^{2} θ

= 1– cos^{2} θ

= sin^{2} θ

**Sol: **

Let **a** = x**i** + y**j** + z**k**

** = i** ( 0 – 0) – **j **(0 – z) + **k** (0 – y)

** = **– y**k **+ z**j**

Similarly, ** ** = x^{2} + z^{2}** **and** ** = y^{2} + x^{2}** **

** + ****+** ** = y**^{2} + z^{2} + x^{2} + z^{2} + y^{2} + x^{2}** **

= 2(x^{2} + y^{2} +z^{2})

If = 2, = 3 and (**p**, **q**) = , then find

**Sol:**

= 2 × 3 sin

= 2 × 3×1/2

= 3

If 4**i** + **j** + p**k **is parallel to the vector **i** + 2**j** + 3**k**, then find p.

**Sol:**

Given 4**i** + **j** + p**k **is parallel to the vector **i** + 2**j** + 3**k**

⇒ p = 12

Compute **a**× (**b** + **c**) + **b**× (**c** + **a**) + **c**× (**a** + **b**)

**Sol:**

**a**× (**b** + **c**) + **b**× (**c** + **a**) + **c**× (**a** + **b**)

**= a**× **b** +** a**× **c **+ **b**× **c** +** b**× **a **+** c **× **a** +** c **× **b**

**= a**× **b** +** a**× **c **+ **b**× **c** –**a **× **b **–** a** × **c **–** b** ×**c**

**= 0**

Compute 2**j**× (3**i – **4**k**) + (**i** + 2**j**) × **k**

**Sol:**

2**j**× (3**i – **4**k**) + (**i** + 2**j**) × **k **

**= **6(**j** × **i**) – 8(**j** × **k**) + (**i **× **k**) + 2(**j** × **k**)

= **– **6**k – **8**i** **– j **+ 2**i**

= **– **6**i –j** **–**6** j**

]]>

**Addition of Vectors**

Find the unit vector in the direction of

The unit vector in the direction of a vector is given by

Find a vector in the direction of **a where** that has a magnitude of 7 units.

The unit vector in the direction of a vector is

The vector having the magnitude 7 and in the direction of is

Find the unit vector in the direction of the sum of the vectors, **a** = 2**i **+ 2**j **– 5**k **and** b **= 2**i **+** j **+ 3**k**

**Sol** Given vectors are **a** = 2**i **+ 2**j **– 5**k **and** b **= 2**i **+** j **+ 3**k**

**a** + **b** = (2**i **+ 2**j **– 5**k) **+ (2**i **+** j **+ 3**k**) = 4**i **+ 3**j **– 2**k**

Write the direction cosines of the vector

**QUESTION 5**

Show that the points whose position vectors are – 2**a **+ 3**b **+ 5**c**, **a **+ 2**b **+ 3**c**, 7** a** – **c** are collinear when **a**,** b**,** c **are non-collinear vectors

**Sol:** Let **OA **= – 2**a **+ 3**b **+ 5**c, OB = a **+ 2**b **+ 3**c**, **OC **= 7** a** – **c****A**

**B **= **OB – OA = a **+ 2**b **+ 3**c **– **(**– 2**a **+ 3**b **+ 5**c)**

** AB = **3**a **–** b **– 2**c**

** AC **= **OC – OA **= 7** a** – **c** – **(**– 2**a **+ 3**b **+ 5**c)**

** AC = **9**a **– 3**b **– 6**c =** 3**(**3**a **–** b **– 2**c)**

** AC = **3 **AB**

** **A, B and C are collinear

ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) **AL** and **AM** in terms of** AB** and **AD (**ii) 𝛌,** if AM = **𝛌 **AD – LM**

**Sol:** Given, ABCD is a parallelogram and L and M are middle points of BC and CD

(i) Take A as the origin

M is the midpoint of CD

** = AD **+ ½** AB **(∵ AB = DC)

L is the midpoint of BC

** = AB **+ ½ **AD **((∵ BC = AD)

(ii) **AM = **𝛌 **AD – LM**

**AM + LM= **𝛌 **AD **

** AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD **

**AD + ½ AB + AD + ½ AB – AB – ½ AD = 𝛌 AD **

3/2 **AD = 𝛌 AD **

∴**𝛌 = 3/2**

If G is the centroid of the triangle ABC, then show that **OG = ** when, are the position vectors of the vertices of triangle ABC.

**Sol:** **OA** = **a, OB **=** b, OC **=** c and OD **=** d**

** **D is the midpoint of BC

G divides median AD in the ratio 2: 1

If = , = are collinear vectors, then find m and n.

**Sol:** Given , are collinear vectors

Equating like vectors

2 = 4 λ; 5 = m λ; 1 = n λ

∴ m = 10, n = 2

Let If , . Find the unit vector in the direction of **a + b.**

** **The unit vector in the direction of **a + b **=

If the vectors – 3**i** + 4**j** + λ**k** and μ**i** + 8**j** + 6**k**. are collinear vectors, then find λ and μ.

**Sol:** let **a **= – 3**i** + 4**j** + λ**k**, **b** = μ**i** + 8**j** + 6**k**

** ⟹ **** a **= t**b**

– 3**i** + 4**j** + λ**k** = t (μ**i** + 8**j** + 6**k)**

– 3**i** + 4**j** + λ**k** = μt **i + **8t **j + **6t **k**

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

∴ μ=– 6, λ = 3

ABCD is a pentagon. If the sum of the vectors **AB**, **AE**, **BC**, **DC**, **ED** and **AC **is 𝛌** AC **then find the value of 𝛌

**Sol:** Given, ABCD is a pentagon

**AB** + **AE** + **BC** + + **DC** + **ED** + **AC **= 𝛌** AC**

(**AB** +** BC) **+ (**AE** + **DC** + **ED**) + **AC = **𝛌** AC**

** AC** + **AC** + **AC = **𝛌** AC**

** 3 AC = **𝛌** AC**

** **𝛌** = **3

If the position vectors of the points A, B and C are – 2**i** + **j** – **k** and –4**i** + 2**j** + 2**k **and 6**i** – 3**j** – 13**k **respectively and** AB = **𝛌** AC, **then find the value of 𝛌

**Sol:** Given, OA = – 2**i** + **j** – **k **, OB = –4**i** + 2**j** + 2**k **and OC = 6**i** – 3**j** – 13**k **

AB = OB – OA = –4**i** + 2**j** + 2**k **– (– 2**i** + **j** – **k**)

= –4**i** + 2**j** + 2**k **+2**i** – **j** + **k**

= –2**i** + **j** + 3**k **

AC = OC – OA = 6**i** – 3**j** – 13**k **– (– 2**i** + **j** – **k**)

= 6**i** – 3**j** – 13**k **+2**i** – **j** + **k**

= 8**i** –4 **j** –12**k**

** = **– 4 (2**i** + **j** + 3**k)**

**AC = **– 4** AB**

Given** AB = **𝛌** AC**

** **𝛌 = – 1/4

If **OA** = **i** + **j** +**k,** **AB** = 3**i** – 2**j** + **k**, BC = **i** + 2**j** – 2**k, CD** = 2**i** + **j** +3**k** then find the vector **OD**

**Sol:** Given **OA** = **i** + **j** +**k,** **AB** = 3**i** – 2**j** + **k**, BC = **i** + 2**j** – 2**k, CD** = 2**i** + **j** +3**k**

** OD = OA **+** AB **+** BC **+** CD **

** = i** + **j** +**k **+ 3**i** – 2**j** + **k** + **i** + 2**j** – 2**k **+ 2**i** + **j** +3**k**

** = **7**i** + 2**j** +4**k**

Let **a** = 2**i** +4 **j** –5 **k**, **b** = **i** + **j+** **k**, **c** = **j** +2 **k**, then find the unit vector in the opposite direction of **a **+ **b** + **c**

**Sol:** Given, **a** = 2**i** +4 **j** –5 **k**, **b** = **i** + **j+** **k**, **c** =** j** +2 **k**

** a **+ **b** + **c **= 2**i** +4 **j** –5 **k** + **i** + **j+** **k **+ **j **+2 **k**

** = **3**i** +6**j** –2**k**

** **The unit vector in the opposite direction of **a **+ **b** + **c **is

Is the triangle formed by the vectors 3**i** +5**j** +2**k**, **2i** –3**j** –5**k**, **–**5**i** – 2**j** +3**k **

**Sol:** Let **a** =3**i** +5**j** +2**k**, **b** = **2i** –3**j** –5**k**,** c** = **–**5**i** – 2**j** +3**k**

∴ Given vectors form an equilateral triangle.

Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are **a**, **b,** (3**a** – 2**b**) are collinear.

**Sol:** the vector equation of the straight line passing through two points** a**, **b **is

**r** = (1 – t) **a**+ t **b **

** **3**a – **2**b = (**1 – t) **a**+ t **b **

** **Equating like vectors

1 – t = 3 and t = – 2

∴ Given points are collinear.

OABC is a parallelogram If OA = **a** and OC = **c**, then find the vector equation of the side **BC**

**Sol:** Given, OABC is a parallelogram and OA = **a**, OC = **c**

The vector equation of** BC is** a line which is passing through C(**c**) and parallel to **OA**

⟹ the vector equation of **BC** is r = **c** + t **a**

If **a**, **b**, **c** are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A

**Sol:** Given **OA** = **a**, **OB** = **b**, **OC** = c

D is mid of BC

The equation of AD is

Find the vector equation of the line passing through the point 2**i** +3**j** +**k **and parallel to the vector 4**i – **2**j **+ 3**k**

Sol: Let **a =**2**i** +3**j** +**k**, **b **= 4**i – **2**j **+ 3**k **

The vector equation of the line passing through **a **and parallel to the vector **b **is **r** = **a **+ t**b**

** r** = 2**i** +3**j** +**k **+ t (4**i – **2**j **+ 3**k)**

= (2 + 4t) **i **+ (3 – 2t) **j** + (1 + 3t)** k**

Find the vector equation of the plane passing through the points **i – **2**j **+ 5**k**, **– **2**j –k **and **– 3i **+ 5**j**

Sol: The vector equation of the line passing through **a,** **b and c**is **r** = (1 – t – s) a + t**b **+ s**c**

** **⟹ **r** = (1 – t – s) (**i – **2**j **+ 5**k**) + t (**– **2**j –k**) + s (**– 3i **+ 5**j**)

= (1 – t – 4s)** i** + (– 2 – 3t + 7s)** j** + (5 – 6t – 5s)** k**

]]>

**Matrices**

If A = , then show that A^{2} = –I

∴ A^{2} = –I

If A = , and A^{2} = 0, then find the value of k.

A^{2} = 0

8 + 4k = 0, – 2 – k = 0 and –4 + k^{2} = 0

4k = –8; k = –2; k^{2} = 4

k = –2; k = –2; k = ± 2

∴ k =– 2

Trace of A = 1 – 1 + 1 = 1

If A = , B = and 2X + A = B, then find X.

**Sol:** Given A = , B = and 2X + A = B

2X = B – A

Find the additive inverse of A, If A =

Additive inverse of A = – A

If , then find the values of x, y, z and a.

⟹ x- 1 = 1 – x ; y – 5 = – y ; z = 2 ; 1 + a = 1

⟹ x + x = 1 + 1; y + y = 5; z = 2; a =1– 1

⟹ 2x = 1; 2y = 5; z = 2; a = 0

∴ x = ½ ; y = 5/2; z = 2; a = 0

Construct 3 × 2 matrix whose elements are defined by a_{ij} =

**Sol:**

Let A= _{ }

a_{11} = 1

a_{22} = 2

a_{31} = 0

If A = and B = , do AB and BA exist? If they exist, find them. BA and AB commutative with respect to multiplication.

**Sol:** Given Matrices are A = B =

Order of A = 2 × 3 and Order of B = 3 × 2

AB and BA exist

AB and BA are not Commutative under Multiplication

Define Symmetric and Skew Symmetric Matrices

**Sol:**

Symmetric Matrix: Let A be any square matrix, if A^{T} = A, then A is called Symmetric Matrix

Skew Symmetric Matrix: Let A be any square matrix if A^{T} = –A, then A is called Skew Symmetric Matrix

If A = is a symmetric matrix, then find x.

**Sol:** Given, A = is a symmetric matrix

⟹ A^{T} = A

⟹ x = 6

If A = is a skew-symmetric matrix, then find x

**Sol:** Given A = is a skew-symmetric matrix

⟹ A^{T} = – A

⟹ x = –x

x+ x = 0 ⟹ 2x = 0

⟹ x = 0

If A = and B = , then find (A B^{T})^{ T}

If A = and B = , then find A + B^{T}

If A = , then show that AA^{T} = A^{T}A = I

∴ AA^{T} = A^{T}A = I

Find the minor of – 1 and 3 in the matrix

**Sol:** Given Matrix is

Find the cofactors 0f 2, – 5 in the matrix

**Sol:** Given matrix is

Cofactor of 2 = (–1)^{2 + 2} = –3 + 20 = 17

Cofactor of – 5 = (–1)^{3 + 2} = –1(2 – 5) = –1(–3) = 3

If ω is a complex cube root of unity, then show that = 0(where 1 + ω+ω^{2} = 0)

R_{1} → R_{1} + R_{2} + R_{3}

If A = and det A = 45, then find x.

Det A = 45

⟹ 1(3x + 24) – 0 (2x – 20) + 0 (– 12 – 15) = 45

⟹ 3x + 24 = 45

3x = 45 – 24

3x = 21

x = 7

Find the adjoint and inverse of the following matrices

(i)

(ii)

Det A = a (bc – 0) – 0(0 – 0) + 0(0 – 0)

Det A = abc ≠ 0

Adj A = (Cofactor matrix of A)^{ T}

Find the rank of the following matrices.

Det A = 1 (0 – 2) – 2(1 – 0) + 1(– 1 – 0)

= – 2– 2– 1

= – 5 ≠ 0

∴ Rank of A = 3

Det A = – 1 (24 – 25) + 2(18 – 20) + – 3(15 – 16)

= – 1– 4 + 3

= – 0

Det B = – 4 + 6 = 2 ≠ 0

∴ Rank of A = 2

Det of Sub matrix of A = – 1 – 0 = – 1 ≠ 0

∴ Rank of A = 2

Det of Sub matrix of A =1 (1 – 0) – 0(0 – 0) + 0(0 – 0)

= 1≠ 0

∴ Rank of A = 3

]]>

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.

Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, – 5).

Sol:

The plane passes through A and perpendicular to OA, then the line segment OA is normal to the plane.

Dr’s of OA = (2, 3, – 5)

The equation of the plane passing through point (x_{1}, y_{1}, z_{1}) and dr’s (a, b, c) is

a(x – x_{1}) + b (y – y_{1}) + c (z – z_{1}) = 0

⟹ 2(x – 2) + 3 (y – 3) – 5 (z + 5) = 0

2x – 4 + 3y – 9 – 5z – 25 = 0

2x + 3y – 5z – 38 = 0

Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4)

Sol:

The equation of the plane passing through the points (x_{1}, y_{1}, z_{1}) (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}) is

The plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4) is

x (30 – 20) – (y + 1) (20 – 6) + (z + 1) (40 – 18) = 0

x (10) – (y + 1) (14) + (z + 1) (22) = 0

10x – 14y – 14 + 22z + 22 = 0

10x – 14y + 22z + 8 = 0

2(5x – 7y + 11z + 4) = 0

∴ the equation of the plane is 5x – 7y + 11z + 4 = 0

Find the equation to the plane parallel to the ZX-plane and passing through (0, 4, 4).

Sol:

Equation of ZX-plane is y = 0

The equation of the plane parallel to the ZX-plane is y = k

But it is passing through (0, 4, 4)

⟹ y = 4

Find the equation to the plane passing through the point (α, β, γ) and parallel to the plane axe + by + cz + d = 0.

Sol:

The equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0

But it is passing through the point (α, β, γ)

a α + b β + c γ + k = 0

⟹ k = – a α – b β – c γ

The equation of the plane is ax + by + cz – a α – b β – c γ = 0

⟹ a(x – α) + b (y – β)+ c (z – γ) = 0

Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 7.

Sol: If θ is the angle between the planes a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 and a_{2}x + b_{2} y + c_{2} z + d_{2} = 0, then cos θ =

Cos θ = cos 60^{0}

θ = 60^{0} =

Reduce the equation x + 2y – 2z – 9 = 0 to the normal form and hence find the dc’s of the normal to the plane.

Sol: Given plane is x + 2y – 2z – 9 = 0

x + 2y – 2z = 9

dc’s of the normal to the plane are

Suppose a plane makes intercepts 2, 3, 4 on X, Y, Z axes respectively. Find the equation of the plane in the intercept form.

Sol: Given a = 2, b = 3, c = 4

The equation of the line in the intercept form is

Express x – 3y + 2z = 9 in the intercept form

Sol: Given plane is x – 3y + 2z = 9

a = 9, b = – 3, c = 9/2

Find the direction cosine of the normal to the plane x + 2y + 2z – 4 = 0.

Sol: Given plane is x + 2y + 2z – 4 = 0

We know that Dr’s of the normal to the plane ax + by + cz + d = 0 are (a, b, c)

⟹ dc’s of the normal to the plane =

⟹ dr’s of the normal to the plane x + 2y + 2z – 4 = 0 are (1, 2, 2)

⟹ dc’s of the normal to the plane are

Find the midpoint of the line joining the points (1, 2, 3) and (–2, 4, 2)

Sol: Given points are A (1, 2, 3), B (–2, 4, 2)

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

**1.Show that the points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) forma a right-angled isosceles triangle.**

**Sol:**

Distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is PQ =

∴ The points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle.

**2.Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 = 0**

**Sol:**

Let P (x, y, z) be the locus of the point

A (0, y, 0) be any point on Y – axis

B = (1, 2, – 1)

Condition is PA = 3PB

PA^{2} = (3PB)^{2}

PA^{2}= 9 PB^{2 }

⟹ x^{2} + z^{2} = 9[(x – 1)^{2} + (y – 2)^{2} + (z + 1)^{2}]

x^{2} + z^{2} = 9[x^{2 }– 2x + 1 + y^{2} – 4y + 4 + z^{2} + 2z + 1]

x^{2} + z^{2} = 9x^{2 }– 18x + 9 + 9y^{2} – 36y + 36 +9 z^{2} + 18z + 9

∴ 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 = 0

**3.A, B, C are three points on OX, OY and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C and O**

**Sol:**

Let P (x, y, z) be the required point

O = (0, 0, 0) A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Given, AP = BP = CP = OP

AP = OP

⟹ AP^{2} =OP^{2}

(x – a )^{2 }+ y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

x^{2} + a^{2} – 2ax + y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

a^{2} – 2ax = 0

a (a – 2x) = 0

a – 2x = 0 (∵ a≠0)

a = 2x ⟹ a/2

Similarly, y = b/2 and z = c/2

∴ P = (a/2, b/2, c/2)

**4.Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear**

**Sol:**

Given points are A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2)

**5.Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.**

**Sol:**

Let A = (5, – 1, 7), B = (x, 5, 1)

Given AB = 9

⟹ AB^{2} = 81

(5 – x)^{2} + (– 1 – 5)^{2} + (7 – 1)^{2} = 81

(5 – x)^{2 }+ 36 + 36 = 81

(5 – x)^{2} + 72 = 81

(5 – x)^{2 }= 81 – 72 = 9

(5 – x) = ± 3

5 – x = 3 or 5 – x = – 3

5 – 3 = x or 5 + 3 = x

x = 2 or x = 8

**6.If the point (1, 2, 3) is changes to the point (2, 3, 1) through translation of axes. Find new origin.**

Sol:

Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)

x = X + h, y = Y + k, z = Z + l

h = x – X, k = y – Y, l = z – Z

h = 1 – 2, k = 2 – 3, l = 3 – 1

h = – 1, k = – 1, l = 2

New origin is (– 1, – 1, 2)

**7.By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.**

**Sol:**

If a point P divides the line segment joining the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) in the ratio, then

Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3

**8.Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).**

**Sol:**

Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)

Suppose P divides AB in the ratio k : 1

**9.Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not**.

**Sol:**

Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)

Let C divides AB in the ratio k : 1

2 – 4k = 4 (k + 1)

2 – 4k = 4k + 4

– 4k– 4k = 4 – 2

– 8k = 2

K = -1/4

C divides AB in the Ratio 1 : 4 externally

∴ A, B, C are collinear

**10.Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)**

**Sol:**

The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is

**11. Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)**

**Sol:**

The centroid of the tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is

**12. Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)**

**Sol:**

Let P be any point on the YZ-plane

P = (0, y, z)

Let P divides AB int eh ratio k:1

YZ-plane divides AB in the ratio – 2:3

**13. Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).**

**Sol: **

let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)

ABCD is a parallelogram

Midpoint of AC = Midpoint of BD

⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0

x = 3, y = 3, z = 1

∴ The fourth vertex D = (3, 3, 1)

**14. A (5, 4, 6), B (1, –1, 3), C (3, 3, 1) are three points. Find the coordinates of the point in which the bisector of ∠BAC meets the side BC.**

**Sol:**

We know that the bisector of ∠BAC divides BC in the ratio AB:AC

If D is a point where the bisector of ∠BAC meets BC

⟹ D divides BC in the ratio 5:3

**15. If M (α, β, γ) is the midpoint of the line joining the points (x _{1}, y_{1}, z_{1}) and B, then find B**

**Sol:**

Let B (x, y, z) be the required point

M is the midpoint of AB

⟹ (α, β, γ) =

⟹ 2 α = x + x_{1}; 2 β = y + y_{1}; 2 γ = z + z_{1}

x =2 α – x_{1}; x =2 β – y_{1;} x =2 γ – z_{1}

_{ }∴ B = (2 α – x_{1}, 2 β – y_{1}, 2 γ –)

**16. If H, G, S and I respectively denote orthocenter, centroid, circumcenter and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.**

**Sol:**

Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)

⟹ ∆ ABC is an equilateral triangle

We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same

Centroid G =

= (2, 2, 2)

∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)

**17. Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).**

**Sol:**

Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)

∴ I = (1, 1,0)

**18. Find the ratio in which the point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also find the hormonic conjugate of P**

**Sol:**

Hormonic Conjugate: If P divides AB in the ratio m : n, then the Hormonic Conjugate of P (i.e., Q) divides AB in the ratio –m : n.

Given points are A (3, 2, – 4), B (9, 8, –10) and P (5, 4, – 6)

P divides AB in the ratio is x_{2} – x : x – x_{2}

= 3 – 5 : 5 – 9

= 1 : 2 (internally)

Let Q be the hormonic conjugate of P

⟹ Q divides AB in the ratio –1 : 2

Q =

= (–3, –4, –2)

Q (–3, –4, –2) is the hormonic conjugate of P (5, 4, – 6)

**19. If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of tetrahedron, find the fourth Vertex.**

**Sol:**

Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)

Let the fourth vertex is D = (x, y, z)

The centroid of the tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

13 + x = 16, 5 + y = 8, 5 + z = 2

x = 3, y = 3, z = 3

∴ the fourth vertex D = (3, 3, 3)

**20. Show that the points A(3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear and find the ratio in which B divides AC.**

**Sol:**

Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)

∴ the points A (3, 2, –4), (5, 4, –6) and C (9, 8, –10) are collinear

B divides AB in the ratio is AB:BC =

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